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Question 1:

Answer:

According to the right-hand thumb rule, if the thumb of our right hand points in the direction of the current flowing, then the curling of the fingers will show the direction of the magnetic field developed due to it and vice versa. Let us consider the case where an electric current flows north to south in a wire.
According to the right-hand thumb rule,
(a) for any point in the east of the wire, the magnetic field will come out of the plane of paper
(b) for a point in the west of the wire, the magnetic field will enter the plane of paper
(c) for any point vertically above the wire, the magnetic field will be from right to left
(d) for any point vertically below the wire, the magnetic field will be from left to right

Page No 248:

Question 2:

According to the right-hand thumb rule, if the thumb of our right hand points in the direction of the current flowing, then the curling of the fingers will show the direction of the magnetic field developed due to it and vice versa. Let us consider the case where an electric current flows north to south in a wire.
According to the right-hand thumb rule,
(a) for any point in the east of the wire, the magnetic field will come out of the plane of paper
(b) for a point in the west of the wire, the magnetic field will enter the plane of paper
(c) for any point vertically above the wire, the magnetic field will be from right to left
(d) for any point vertically below the wire, the magnetic field will be from left to right

Answer:

The magnetic field due to a long, straight wire is given by
B = μoi2πdSpeed of light, c = 1μoεoμo = 1c2εoB = i2πc2εod
(In terms of ε0ci and d)

Page No 248:

Question 3:

The magnetic field due to a long, straight wire is given by
B = μoi2πdSpeed of light, c = 1μoεoμo = 1c2εoB = i2πc2εod
(In terms of ε0ci and d)

Answer:

According to the right-hand thumb rule, if we curl the fingers of our right hand in the direction of the current flowing, then the thumb will point in the direction of the magnetic field developed due to it and vice versa. Therefore, in this case, the field at the centre is going away from us.

Page No 248:

Question 4:

According to the right-hand thumb rule, if we curl the fingers of our right hand in the direction of the current flowing, then the thumb will point in the direction of the magnetic field developed due to it and vice versa. Therefore, in this case, the field at the centre is going away from us.

Answer:

In Ampere's law B.dl  = μoii is the total current crossing the area bounded by the closed curve. The magnetic field B on the left-hand side is the resultant field due to all existing currents.  

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Question 5:

In Ampere's law B.dl  = μoii is the total current crossing the area bounded by the closed curve. The magnetic field B on the left-hand side is the resultant field due to all existing currents.  

Answer:

The magnetic field due to a long solenoid is given as B = µ0ni, where n is the number of loops per unit length. So, if we add more loops at the ends of the solenoid, there will be an increase in the number of loops and an increase in the length, due to which the ratio n will remain unvaried, thereby leading to not a considerable effect on the field inside the solenoid.

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Question 6:

The magnetic field due to a long solenoid is given as B = µ0ni, where n is the number of loops per unit length. So, if we add more loops at the ends of the solenoid, there will be an increase in the number of loops and an increase in the length, due to which the ratio n will remain unvaried, thereby leading to not a considerable effect on the field inside the solenoid.

Answer:

Ampere's law is valid for a loop that is not circular. However, it should have some charge distribution in the area enclosed so as to have a constant electric field in the region and a non-zero magnetic field. Even if the loop defined does not have its own charge distribution but has electric influence of some other charge distribution, it can have some constant magnetic field (B.dl = μoienclosed).

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Question 7:

Ampere's law is valid for a loop that is not circular. However, it should have some charge distribution in the area enclosed so as to have a constant electric field in the region and a non-zero magnetic field. Even if the loop defined does not have its own charge distribution but has electric influence of some other charge distribution, it can have some constant magnetic field (B.dl = μoienclosed).

Answer:

The magnetic force on a wire carrying an electric current i is F=i.(l×B), where l is the length of the wire and B is the magnetic field acting on it. If a uniformly charged ring starts rotating around a straight wire, then according to the right-hand thumb rule, the magnetic field due to the ring on the current carrying straight wire placed at its axis will be parallel to it. So, the cross product will be
(l×B)=0F=0
Therefore, no magnetic force will act on the wire.

Page No 248:

Question 8:

The magnetic force on a wire carrying an electric current i is F=i.(l×B), where l is the length of the wire and B is the magnetic field acting on it. If a uniformly charged ring starts rotating around a straight wire, then according to the right-hand thumb rule, the magnetic field due to the ring on the current carrying straight wire placed at its axis will be parallel to it. So, the cross product will be
(l×B)=0F=0
Therefore, no magnetic force will act on the wire.

Answer:

The magnetic force on a wire carrying an electric current i is F=i.(l×B), ​where l is the length of the wire and B is the magnetic field acting on it. Suppose we have one wire in the horizontal direction (fixed) and other wire in the vertical direction (free to move). If the horizontal wire is carrying current from right to left is held fixed perpendicular to the vertical wire, which is free to move, the upper portion of the free wire will tend to move in the left direction and the lower portion of the wire will tend to move in the right direction, according to Fleming's left hand rule, as the magnetic field acting on the wire due to the fixed wire will point into the plane of paper above the wire and come out of the paper below the horizontal wire and the current will flow in upward direction in the free wire. Thus, the free wire will tend to become parallel to the fixed wire so as to experience maximum attractive force.

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Question 9:

The magnetic force on a wire carrying an electric current i is F=i.(l×B), ​where l is the length of the wire and B is the magnetic field acting on it. Suppose we have one wire in the horizontal direction (fixed) and other wire in the vertical direction (free to move). If the horizontal wire is carrying current from right to left is held fixed perpendicular to the vertical wire, which is free to move, the upper portion of the free wire will tend to move in the left direction and the lower portion of the wire will tend to move in the right direction, according to Fleming's left hand rule, as the magnetic field acting on the wire due to the fixed wire will point into the plane of paper above the wire and come out of the paper below the horizontal wire and the current will flow in upward direction in the free wire. Thus, the free wire will tend to become parallel to the fixed wire so as to experience maximum attractive force.

Answer:

Two proton beams going in the same direction repel each other, as they are like charges and we know that like charges repel each other.
When a charge is in motion then a magnetic field is associated with it. Two wires carrying currents in the same direction produce their fields (acting on each other) in opposite directions so the resulting magnetic force acting on them is attractive. Due to the magnetic force, these two wires attract each other.
But when a charge is at rest then only an electric field is associated with it and no magnetic fiels is produced by it. So at rest, it repels a like charge by exerting a electric force on it.
Charge in motion can produce both electric field and magnetic field.
The attractive force between two current carrying wires is due to the magnetic field and repulsive force is due to the electric field.

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Question 10:

Two proton beams going in the same direction repel each other, as they are like charges and we know that like charges repel each other.
When a charge is in motion then a magnetic field is associated with it. Two wires carrying currents in the same direction produce their fields (acting on each other) in opposite directions so the resulting magnetic force acting on them is attractive. Due to the magnetic force, these two wires attract each other.
But when a charge is at rest then only an electric field is associated with it and no magnetic fiels is produced by it. So at rest, it repels a like charge by exerting a electric force on it.
Charge in motion can produce both electric field and magnetic field.
The attractive force between two current carrying wires is due to the magnetic field and repulsive force is due to the electric field.

Answer:

We can obtain a magnetic field due to a straight, long wire using Ampere's law by mentioning the current flowing in the wire, without emphasising on the source of the current in the wire. To apply Ampere's circuital law, we need to have a constant current flowing in the wire, irrespective of its source.

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Question 11:

We can obtain a magnetic field due to a straight, long wire using Ampere's law by mentioning the current flowing in the wire, without emphasising on the source of the current in the wire. To apply Ampere's circuital law, we need to have a constant current flowing in the wire, irrespective of its source.

Answer:

Connecting wires carrying currents in opposite directions are twisted together in using electrical appliances.If the wire is twisted, then the resultant fields from consecutive twists are in opposite directions. So the cumulative effect over a long length of wire is roughly zero.

Let at any point in between the two wires, B1 and B2 be the magnetic field due to wire 1 and wire 2 respectively.

From the diagram, we can see that the net magnetic field due to first turn is into the paper and due to second twisted turn is out of the plane of paper so these fields will cancel each other. Hence if the wire is twisted, then the resultant fields from consecutive twists are in opposite directions. So the cumulative effect over a long length of wire is roughly zero.

Page No 248:

Question 12:

Connecting wires carrying currents in opposite directions are twisted together in using electrical appliances.If the wire is twisted, then the resultant fields from consecutive twists are in opposite directions. So the cumulative effect over a long length of wire is roughly zero.

Let at any point in between the two wires, B1 and B2 be the magnetic field due to wire 1 and wire 2 respectively.

From the diagram, we can see that the net magnetic field due to first turn is into the paper and due to second twisted turn is out of the plane of paper so these fields will cancel each other. Hence if the wire is twisted, then the resultant fields from consecutive twists are in opposite directions. So the cumulative effect over a long length of wire is roughly zero.

Answer:

Magnetic field can not do any work and hence can never speed up or down a particle.
Consider 2 wires carrying current in upward direction.
Magnetic field due to current in wire 1 produces a magnetic field out of the plane of paper at the position of wire 2. Due to this magnetic field, a force is exerted on wire 2. Wire 2 electron, moving in downward direction,  move in circular paths due to this magnetic force. As these electrons can not come out of the wire so while describing circular path,they hit the edges of the wire and tranfer a momentum to the wire. Due to this change in momentum, wire starts moving and gains kinetic energy.

Page No 248:

Question 1:

Magnetic field can not do any work and hence can never speed up or down a particle.
Consider 2 wires carrying current in upward direction.
Magnetic field due to current in wire 1 produces a magnetic field out of the plane of paper at the position of wire 2. Due to this magnetic field, a force is exerted on wire 2. Wire 2 electron, moving in downward direction,  move in circular paths due to this magnetic force. As these electrons can not come out of the wire so while describing circular path,they hit the edges of the wire and tranfer a momentum to the wire. Due to this change in momentum, wire starts moving and gains kinetic energy.

Answer:

(c) upwards

A vertical wire is carrying current in upward direction, so the magnetic field produced will be anticlockwise (according to the right-hand thumb rule). As the electron beam is sent horizontally towards the wire, the direction of the current will be horizontally away from the wire (direction of conventional current is opposite to the direction of the negative charge). According to Fleming's left-hand rule, the force will act in upward direction, deflecting the beam in the same direction.

Page No 248:

Question 2:

(c) upwards

A vertical wire is carrying current in upward direction, so the magnetic field produced will be anticlockwise (according to the right-hand thumb rule). As the electron beam is sent horizontally towards the wire, the direction of the current will be horizontally away from the wire (direction of conventional current is opposite to the direction of the negative charge). According to Fleming's left-hand rule, the force will act in upward direction, deflecting the beam in the same direction.

Answer:

(c) will not exert any force on the circular loop

The magnetic force on a wire carrying an electric current i is given as​ F=i.(l×B), where l is the length of the wire and B is the magnetic field acting on it. If a current-carrying straight wire is kept along the axis of a circular loop carrying current, then according to the right-hand thumb rule, the magnetic field due to the wire on the current-carrying loop will be along its circumference, which contains a current element i dl.
So, the cross product will be​
(l×B)=0F=0
Thus, the straight wire will not exert any force on the loop.

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Question 3:

(c) will not exert any force on the circular loop

The magnetic force on a wire carrying an electric current i is given as​ F=i.(l×B), where l is the length of the wire and B is the magnetic field acting on it. If a current-carrying straight wire is kept along the axis of a circular loop carrying current, then according to the right-hand thumb rule, the magnetic field due to the wire on the current-carrying loop will be along its circumference, which contains a current element i dl.
So, the cross product will be​
(l×B)=0F=0
Thus, the straight wire will not exert any force on the loop.

Answer:

(a) towards the proton beam
A proton beam is going from north to south, so the direction of the current due to the beam will also be from north to south. Also, an electron beam is going from south to north, so the direction of the current due to the beam will also be  from north to south. The direction of conventional current is along the direction of the flow of the positive charge and opposite to the flow of the negative charge. The magnetic field generated due to them will enter the plane of paper in the west and come out of the plane of paper in the east, according to the right-hand thumb rule. Since both the beams have currents in the same direction, they will apply equal and opposite forces on each other and, hence, will attract each other. Thus, the electron beam will be deflected towards the proton beam.

Page No 248:

Question 4:

(a) towards the proton beam
A proton beam is going from north to south, so the direction of the current due to the beam will also be from north to south. Also, an electron beam is going from south to north, so the direction of the current due to the beam will also be  from north to south. The direction of conventional current is along the direction of the flow of the positive charge and opposite to the flow of the negative charge. The magnetic field generated due to them will enter the plane of paper in the west and come out of the plane of paper in the east, according to the right-hand thumb rule. Since both the beams have currents in the same direction, they will apply equal and opposite forces on each other and, hence, will attract each other. Thus, the electron beam will be deflected towards the proton beam.

Answer:

(d) is towards west at both A and B
According to the right-hand thumb rule, if we curl the fingers of our right hand in the direction of the current flowing, then the stretching of the thumb will show the direction of the magnetic field developed due to it and vice versa.
Let north-south is along x axis and east-west is along y axis. Circular wire is in xz plane. Then point A will lie on positive y axis and B on negative y axis. On looking from point B, current is flowing in anticlockwise direction so the magnetic field will point from right to left. Hence, the magnetic field due to the loop will be towards west at both A and B.

Page No 248:

Question 5:

(d) is towards west at both A and B
According to the right-hand thumb rule, if we curl the fingers of our right hand in the direction of the current flowing, then the stretching of the thumb will show the direction of the magnetic field developed due to it and vice versa.
Let north-south is along x axis and east-west is along y axis. Circular wire is in xz plane. Then point A will lie on positive y axis and B on negative y axis. On looking from point B, current is flowing in anticlockwise direction so the magnetic field will point from right to left. Hence, the magnetic field due to the loop will be towards west at both A and B.

Answer:

(b) move towards the wire


Force acting on the wire per unit length carrying current i2 due to the wire carrying current i1 placed at a distance d is given by
F = μoi1i22πd
So, forces per unit length acting on sides AB and CD are as follows:
FAB = μoi1i22πd    (Towards the wire)FCD = μoi1i22π(d+a)  (Away from the wire)
Here, FAB > FCD  because force is inversly proportional to the distance from the wire and wire AB is closer to the wire carrying current i1.
The forces per unit length acting on sides BC and DA will be equal and opposite, as they are equally away from the wire carrying current i1, with current i2 flowing in the opposite direction.
FBC -FDA
Now,
Net force:
F = FAB+FBC+FCD+FDAF = μoi1i22πd+FBC-μoi1i22π(d+a)-FBCF = μoi1i22π1d-1d+aF = μoi1i2a2πd(d+a)

(Towards the wire)
Therefore, the loop will move towards the wire.

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Question 6:

(b) move towards the wire


Force acting on the wire per unit length carrying current i2 due to the wire carrying current i1 placed at a distance d is given by
F = μoi1i22πd
So, forces per unit length acting on sides AB and CD are as follows:
FAB = μoi1i22πd    (Towards the wire)FCD = μoi1i22π(d+a)  (Away from the wire)
Here, FAB > FCD  because force is inversly proportional to the distance from the wire and wire AB is closer to the wire carrying current i1.
The forces per unit length acting on sides BC and DA will be equal and opposite, as they are equally away from the wire carrying current i1, with current i2 flowing in the opposite direction.
FBC -FDA
Now,
Net force:
F = FAB+FBC+FCD+FDAF = μoi1i22πd+FBC-μoi1i22π(d+a)-FBCF = μoi1i22π1d-1d+aF = μoi1i2a2πd(d+a)

(Towards the wire)
Therefore, the loop will move towards the wire.

Answer:

(d) zero 

The force on a charged particle q moving with velocity v in a magnetic field B is given by
F=q(v×B)
As the charge is moving along the magnetic line of force, the velocity and magnetic field vectors will point in the same direction, making a cross product.
(v×B) = 0F = 0
So, the magnetic force on the particle will be zero.

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Question 7:

(d) zero 

The force on a charged particle q moving with velocity v in a magnetic field B is given by
F=q(v×B)
As the charge is moving along the magnetic line of force, the velocity and magnetic field vectors will point in the same direction, making a cross product.
(v×B) = 0F = 0
So, the magnetic force on the particle will be zero.

Answer:

(c) both of them

Because of the presence of a charge, a particle produces an electric field. Also, because of its motion, that is, the flow of charge or current, there is generation of a magnetic field.



Page No 249:

Question 8:

(c) both of them

Because of the presence of a charge, a particle produces an electric field. Also, because of its motion, that is, the flow of charge or current, there is generation of a magnetic field.

Answer:

(c) the kinetic energy

When a particle of mass m carrying charge q is projected with speed v in a plane perpendicular to a uniform magnetic field B, the field tends to deflect the particle in a circular path of radius r
mv2r = qvBr = mvqBNow,Area, A = πr2A = πmvqB2A = kv2Here, k = πmqB2
Kinetic energy of the particle, E = 12mv2
Therefore, the area bounded is proportional to the kinetic energy.

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Question 9:

(c) the kinetic energy

When a particle of mass m carrying charge q is projected with speed v in a plane perpendicular to a uniform magnetic field B, the field tends to deflect the particle in a circular path of radius r
mv2r = qvBr = mvqBNow,Area, A = πr2A = πmvqB2A = kv2Here, k = πmqB2
Kinetic energy of the particle, E = 12mv2
Therefore, the area bounded is proportional to the kinetic energy.

Answer:

(c)R12R22
Particles X and Y of respective masses m1 and m2 are carrying charge q describing circular paths with respective radii R1 and R2 such that
R1 = m1v1qBR2 = m2v2qB
Since both the particles are accelerated through the same potential difference, both will have the same kinetic energy.
12m1v12 = 12m2v22R1=m1v1qBv1 = R1qBm1And,R2 = m2v2qB v2 = R2qBm2m1R1qBm12 = m2R2qBm22m1m2 = R12R22

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Question 10:

(c)R12R22
Particles X and Y of respective masses m1 and m2 are carrying charge q describing circular paths with respective radii R1 and R2 such that
R1 = m1v1qBR2 = m2v2qB
Since both the particles are accelerated through the same potential difference, both will have the same kinetic energy.
12m1v12 = 12m2v22R1=m1v1qBv1 = R1qBm1And,R2 = m2v2qB v2 = R2qBm2m1R1qBm12 = m2R2qBm22m1m2 = R12R22

Answer:

(b) towards 40 A

​According to Fleming's left-hand rule, if the forefinger and middle finger of our left hand point towards the magnetic field acting on a wire and the current flowing in the wire, respectively, then the thumb will point towards the direction in which the force will act (keeping all three perpendicular). Direction of force can be determined using Fleming's left-hand rule.

In the figure, dotted circle shows the magnetic filed lines due to both the wires.
Magnetic field at any point on the middle wire will be acting along the tangent to the masgnetic field lines at that point.
Therefore, the wire will experience a magnetic field pointing ​towards the 40 A wire.
Due to AB, the force will be towards right and due to CD, the force on the wire will be towards right. So, both the forces will add to give a resultant force, which will be towards right, that is, towards the 40 A current-carrying wire.

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Question 11:

(b) towards 40 A

​According to Fleming's left-hand rule, if the forefinger and middle finger of our left hand point towards the magnetic field acting on a wire and the current flowing in the wire, respectively, then the thumb will point towards the direction in which the force will act (keeping all three perpendicular). Direction of force can be determined using Fleming's left-hand rule.

In the figure, dotted circle shows the magnetic filed lines due to both the wires.
Magnetic field at any point on the middle wire will be acting along the tangent to the masgnetic field lines at that point.
Therefore, the wire will experience a magnetic field pointing ​towards the 40 A wire.
Due to AB, the force will be towards right and due to CD, the force on the wire will be towards right. So, both the forces will add to give a resultant force, which will be towards right, that is, towards the 40 A current-carrying wire.

Answer:

(c) 2

The magnetic field due to the current-carrying long, straight wire at point a is given by
B = μoi2πa
When both the wires carry currents i1 and i2 â€‹in the same direction, they produce magnetic fields in opposite directions at any point in between the wires.
B' = μoi12πa-μoi22πa = 10 μT                           ..(1)
Here, a is the distance of the midpoint from both the wires.
When both the wires carry currents in opposite directions, they produce fields in the same direction at the midpoint of the two wires.
B'' = μoi12πa+μoi22πa = 30 μT         ..(2)
On solving eqs. (1) and (2), we get
i1-i2 = 10i1+i2 = 30i1 = 20, i2 = 10i1i2 = 21 = 2

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Question 12:

(c) 2

The magnetic field due to the current-carrying long, straight wire at point a is given by
B = μoi2πa
When both the wires carry currents i1 and i2 â€‹in the same direction, they produce magnetic fields in opposite directions at any point in between the wires.
B' = μoi12πa-μoi22πa = 10 μT                           ..(1)
Here, a is the distance of the midpoint from both the wires.
When both the wires carry currents in opposite directions, they produce fields in the same direction at the midpoint of the two wires.
B'' = μoi12πa+μoi22πa = 30 μT         ..(2)
On solving eqs. (1) and (2), we get
i1-i2 = 10i1+i2 = 30i1 = 20, i2 = 10i1i2 = 21 = 2

Answer:

(a) μoi2πr
Magnetic field will be independent of the motion of the observer because the velocity with which the observer is moving is comparable to drift velocity of electron which is very small as compared to the speed of flow of current from one end of wire to other end. So it can be neglected  and hence, magnetic field due to the wire w.r.t the observer will be
B = μoi2πr
 

μ0i2πr

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Question 1:

(a) μoi2πr
Magnetic field will be independent of the motion of the observer because the velocity with which the observer is moving is comparable to drift velocity of electron which is very small as compared to the speed of flow of current from one end of wire to other end. So it can be neglected  and hence, magnetic field due to the wire w.r.t the observer will be
B = μoi2πr
 

μ0i2πr

Answer:

(a) μ0i4πdl ×r r3
(b) -μ0i4πr ×dl r3

The magnetic field at the origin due to current element i dl placed at a position r is given by
 dB = μo4πir×dlr3 
According to the cross product property,
A×B = -B×AdB=-μo4πir×dlr3

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Question 2:

(a) μ0i4πdl ×r r3
(b) -μ0i4πr ×dl r3

The magnetic field at the origin due to current element i dl placed at a position r is given by
 dB = μo4πir×dlr3 
According to the cross product property,
A×B = -B×AdB=-μo4πir×dlr3

Answer:

(a) x, y have the same dimensions.
(b) y, z have the same dimensions.
(c) z, x have the same dimensions.

Lorentz Force:
qvB = qEDimensions of x = [v] = EB = [LT-1]
y = 1μoεo = 4πμo  ×14πεo = 9×10910-7 = 3×108 = cDimensions of y = [c] = [LT-1]
Time constant of RC circuit = RC so dimensionally [RC] = [T]
z = lRC  [z] = [LT-1]
Therefore, x, y and z have the same dimensions.

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Question 3:

(a) x, y have the same dimensions.
(b) y, z have the same dimensions.
(c) z, x have the same dimensions.

Lorentz Force:
qvB = qEDimensions of x = [v] = EB = [LT-1]
y = 1μoεo = 4πμo  ×14πεo = 9×10910-7 = 3×108 = cDimensions of y = [c] = [LT-1]
Time constant of RC circuit = RC so dimensionally [RC] = [T]
z = lRC  [z] = [LT-1]
Therefore, x, y and z have the same dimensions.

Answer:

(b) the directions of the magnetic fields are the same
(c) the magnitudes of the magnetic fields are equal
(d) the field at one point is opposite to that at the other point.
Consider a current carrying wire lying along x axis.
At any two points on z axis which are at equal distance from the wire,one above the wire and one below the wire,the magnitude of magnetic field will be same and their directions will be opposite to each other.
At any two points on z axis which are at different distances from the wire,one above the wire and other also above the wire,the magnitude of magnetic field will be different and their directions will be same to each other.

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Question 4:

(b) the directions of the magnetic fields are the same
(c) the magnitudes of the magnetic fields are equal
(d) the field at one point is opposite to that at the other point.
Consider a current carrying wire lying along x axis.
At any two points on z axis which are at equal distance from the wire,one above the wire and one below the wire,the magnitude of magnetic field will be same and their directions will be opposite to each other.
At any two points on z axis which are at different distances from the wire,one above the wire and other also above the wire,the magnitude of magnetic field will be different and their directions will be same to each other.

Answer:

(b) minimum at the axis of the wire
(c) maximum at the surface of the wire

A long, straight wire of radius R is carrying current i, which is uniformly distributed over its cross section. According to Ampere's law, 
B.dl = μoiinsideAt surface, B×2πR = μoiBsurface = μoi2πRInside, B×2πr = μoi'  for r<R 
Here i,is the current enclosed by the amperian loop drawn inside the wire.
Binside will be proportional to the distance from the axis.
On the axis
B =0 
The magnetic fields from points on the cross section will point in opposite directions and will cancel each other at the centre.
Outside, B×2πr = μoiBoutside = μoi2πr, r>R
Therefore, the magnitude of the magnetic field is maximum at the surface of the wire and minimum at the axis of the wire.

Page No 249:

Question 5:

(b) minimum at the axis of the wire
(c) maximum at the surface of the wire

A long, straight wire of radius R is carrying current i, which is uniformly distributed over its cross section. According to Ampere's law, 
B.dl = μoiinsideAt surface, B×2πR = μoiBsurface = μoi2πRInside, B×2πr = μoi'  for r<R 
Here i,is the current enclosed by the amperian loop drawn inside the wire.
Binside will be proportional to the distance from the axis.
On the axis
B =0 
The magnetic fields from points on the cross section will point in opposite directions and will cancel each other at the centre.
Outside, B×2πr = μoiBoutside = μoi2πr, r>R
Therefore, the magnitude of the magnetic field is maximum at the surface of the wire and minimum at the axis of the wire.

Answer:

(b) is constant inside the tube
(c) is zero at the axis

A hollow tube is carrying uniform electric current along its length, so the current enclosed inside the tube is zero.
According to Ampere's law, 
B.dl = μoiinsideInside the tube,B.dl = 0, r<RBinside = ConstantBaxis = 0 

The  magnetic fields from points on the circular surface will point in opposite directions and cancel each other.
Outside the tube, B×2πr = μoiBoutside = μoi2πr, r>R

Page No 249:

Question 6:

(b) is constant inside the tube
(c) is zero at the axis

A hollow tube is carrying uniform electric current along its length, so the current enclosed inside the tube is zero.
According to Ampere's law, 
B.dl = μoiinsideInside the tube,B.dl = 0, r<RBinside = ConstantBaxis = 0 

The  magnetic fields from points on the circular surface will point in opposite directions and cancel each other.
Outside the tube, B×2πr = μoiBoutside = μoi2πr, r>R

Answer:

(a) outside the cable
(b) inside the inner conductor
According to Ampere's law, in a coaxial, straight cable carrying currents i in the inner conductor and -i (equally in the opposite direction) in the outside conductor.
Inside the inner conductor
B.dl = μoiinsideB.dl = 0B.l = 0B = 0
In between the 2 conductors
B.dl = μoiB = μoi2πr
Outside the outer conductor
B.dl = μo(i-i)B = 0
Therefore, the magnetic field is zero outside the cable.

Page No 249:

Question 7:

(a) outside the cable
(b) inside the inner conductor
According to Ampere's law, in a coaxial, straight cable carrying currents i in the inner conductor and -i (equally in the opposite direction) in the outside conductor.
Inside the inner conductor
B.dl = μoiinsideB.dl = 0B.l = 0B = 0
In between the 2 conductors
B.dl = μoiB = μoi2πr
Outside the outer conductor
B.dl = μo(i-i)B = 0
Therefore, the magnetic field is zero outside the cable.

Answer:

(b) The magnetic field at the axis of the conductor is zero.
(c) The electric field in the vicinity of the conductor is zero.
As the current is flowing through a conductor so it it is distributed only on the surface of the conductor not in the volume of the cylindrical conductor. It is equivalent to charge distribution on a cylindrical sheet for which electric field inside a conducting cylindrical sheet is always zero.
Magnetic fields at any point inside the conducting cylinder is proportional to the distance from the axis of the cylinder.
At the axis, r = 0. This implies that field will be zero at the axis.

Page No 249:

Question 1:

(b) The magnetic field at the axis of the conductor is zero.
(c) The electric field in the vicinity of the conductor is zero.
As the current is flowing through a conductor so it it is distributed only on the surface of the conductor not in the volume of the cylindrical conductor. It is equivalent to charge distribution on a cylindrical sheet for which electric field inside a conducting cylindrical sheet is always zero.
Magnetic fields at any point inside the conducting cylinder is proportional to the distance from the axis of the cylinder.
At the axis, r = 0. This implies that field will be zero at the axis.

Answer:

Using the relation F =qv ×B, we get
 B=Fqv  =FItv
Units of
Force (F) = N
Current (I) = A
Time (T) = s
Velocity (v) = m/s
B=N/A-m
Now, using the relation B=μ0i2πr, we get
μ0=B2πri=NA-m×mAμ0=N/A2

Page No 249:

Question 2:

Using the relation F =qv ×B, we get
 B=Fqv  =FItv
Units of
Force (F) = N
Current (I) = A
Time (T) = s
Velocity (v) = m/s
B=N/A-m
Now, using the relation B=μ0i2πr, we get
μ0=B2πri=NA-m×mAμ0=N/A2

Answer:

Given:
Magnitude of current, I = 10 A
Separation of the point from the wire, d = 1 m

  

The magnetic field B  at point (1 m, 0, 0) is given by
B=μ0i2πd=4π×10-7×102π×1B=2×10-6 T
(Along the +ve y-axis by the right-hand thumb rule)



Page No 250:

Question 3:

Given:
Magnitude of current, I = 10 A
Separation of the point from the wire, d = 1 m

  

The magnetic field B  at point (1 m, 0, 0) is given by
B=μ0i2πd=4π×10-7×102π×1B=2×10-6 T
(Along the +ve y-axis by the right-hand thumb rule)

Answer:

Given:
Magnitude of current, I = 10 A
Diameter of the wire, d = 1.6 × 10−3 m
∴ Radius of the wire = 0.8 × 10−3 m

The magnetic field intensity is given by
B=μ0i2πr
  =2×10-7×200.8×10-3=5 mT

Page No 250:

Question 4:

Given:
Magnitude of current, I = 10 A
Diameter of the wire, d = 1.6 × 10−3 m
∴ Radius of the wire = 0.8 × 10−3 m

The magnetic field intensity is given by
B=μ0i2πr
  =2×10-7×200.8×10-3=5 mT

Answer:

Given:
Magnitude of current, I = 100 A
Separation of the road from the wire, d = 8 m
Thus, the magnetic field is given by
B=μ0i2πd=2×10-7×1008= 2.5 mT

Page No 250:

Question 5:

Given:
Magnitude of current, I = 100 A
Separation of the road from the wire, d = 8 m
Thus, the magnetic field is given by
B=μ0i2πd=2×10-7×1008= 2.5 mT

Answer:

Given:
Uniform magnetic field, B0 = 1.0 × 10−5 T    (Vertically upwards)
Separation of the point from the wire, d = 2 cm = 0.02 m
The magnetic field due to the wire is given by
Bw = μ0i2πd = 4π×10-7×1 2π×0.02Bw=1×10-5 T


Now,
Net magnetic field at point P:
BPBw+ B0 = 2 × 10−5 T
Net magnetic field at point Q:
BQBw B0 = 0

Page No 250:

Question 6:

Given:
Uniform magnetic field, B0 = 1.0 × 10−5 T    (Vertically upwards)
Separation of the point from the wire, d = 2 cm = 0.02 m
The magnetic field due to the wire is given by
Bw = μ0i2πd = 4π×10-7×1 2π×0.02Bw=1×10-5 T


Now,
Net magnetic field at point P:
BPBw+ B0 = 2 × 10−5 T
Net magnetic field at point Q:
BQBw B0 = 0

Answer:

(a) As the wire in question is carrying current, so it will also generate a magnetic field around it. And for a long straight wire it will be maximum at the mid-point called P.
Now,
Magnetic field generated by the current carrying wire=μoi2πr
Net magnetic field = B+μ0i2πr

(b) Magnetic field B = 0
when r<μ0i2πB
Clearly,
B = 0
when r=μ0i2πB
But when r>μ0i2πB,
Net magnetic field = B-μ0i2πr

Page No 250:

Question 7:

(a) As the wire in question is carrying current, so it will also generate a magnetic field around it. And for a long straight wire it will be maximum at the mid-point called P.
Now,
Magnetic field generated by the current carrying wire=μoi2πr
Net magnetic field = B+μ0i2πr

(b) Magnetic field B = 0
when r<μ0i2πB
Clearly,
B = 0
when r=μ0i2πB
But when r>μ0i2πB,
Net magnetic field = B-μ0i2πr

Answer:

Given:
Uniform magnetic field, B0 = 4.0 × 10−4 T
Magnitude of current, I = 30 A
Separation of the point from the wire, d = 0.02 m
Thus, the magnetic field due to current in the wire is given by
B = μ0I2πd
   
    = 2×10-7×300.02= 3×10-4 T

B0 is perpendicular to B (as shown in the figure).
∴ Resultant magnetic field
Bnet=B2+B02      =(4×10-4)2+(3×10-4)2      =5×10-4 T

Page No 250:

Question 8:

Given:
Uniform magnetic field, B0 = 4.0 × 10−4 T
Magnitude of current, I = 30 A
Separation of the point from the wire, d = 0.02 m
Thus, the magnetic field due to current in the wire is given by
B = μ0I2πd
   
    = 2×10-7×300.02= 3×10-4 T

B0 is perpendicular to B (as shown in the figure).
∴ Resultant magnetic field
Bnet=B2+B02      =(4×10-4)2+(3×10-4)2      =5×10-4 T

Answer:

Given:
Uniform magnetic field, B0 =2.0 × 10−3 T (From south to north)
To make the resultant magnetic field zero, the magnetic field due to the wire should be of the same magnitude as B0 and in the direction north to south.
The above condition will be satisfied when the required point will be placed in the west w.r.t. the wire.
Let the separation of the point from the wire be d.

The magnetic field due to current in the wire is given by
B=μ0I2πd

From the question, B = B0.
2.0×10-3 = μ0I2πd
2.0×10-3=2×10-7×10dd=10-3 m=1 mm

Page No 250:

Question 9:

Given:
Uniform magnetic field, B0 =2.0 × 10−3 T (From south to north)
To make the resultant magnetic field zero, the magnetic field due to the wire should be of the same magnitude as B0 and in the direction north to south.
The above condition will be satisfied when the required point will be placed in the west w.r.t. the wire.
Let the separation of the point from the wire be d.

The magnetic field due to current in the wire is given by
B=μ0I2πd

From the question, B = B0.
2.0×10-3 = μ0I2πd
2.0×10-3=2×10-7×10dd=10-3 m=1 mm

Answer:

For point A1,
Magnitude of current in wires, I = 10 A
Separation of point A1from the wire on the left side, d = 2 cm
Separation of point A1from the wire on the right side, d' = 6 cm



In the figure
Red and blue arrow denotes the direction of magnetic field due to the wire marked as red and blue respectively.
P (marked red) denotes the wire carrying current in a plane going into the paper.
Q (marked blue) denotes the wire carrying current in a plane coming out of the paper.
Also from the figure, we can see that
 PA4 = QA4  A4A3P = A4A3Q = 90° A4PA3 =  A4QA3 = 45° PA4A3 =  QA4A3 = 45° PA4Q = 90°   
The magnetic field at A1 due to current in the wires is given by
B=μ0I2πd-μ0I2πd'    ...(1)

B= 2×10-7×102×10-2-2×10-7×106×10-2        = 1-13×10-4        = 0.67×10-4 T

Similarly, we get the magnetic field at A2 using eq. (1).
B= 2×10-7×101×10-2+2×10-7×103×10-2   =83×10-4 T   =2.67×10-4 T

Now,
Magnetic field at A3:
B = 2×10-7×102×10-2+2×10-7×102×10-2   = 2×10-4 T

Magnetic field at A4:
Separation of point A4from the wire on the left side, d = 22+22 =22 cm
Separation of point A4from the wire on the right side, d' =22+22 =22 cm
Thus, the magnetic field at A4 due to current in the wires is given by
B = 2×10-7×1022×10-22+2×10-7×1022×10-22   =1×10-4 T

Page No 250:

Question 10:

For point A1,
Magnitude of current in wires, I = 10 A
Separation of point A1from the wire on the left side, d = 2 cm
Separation of point A1from the wire on the right side, d' = 6 cm



In the figure
Red and blue arrow denotes the direction of magnetic field due to the wire marked as red and blue respectively.
P (marked red) denotes the wire carrying current in a plane going into the paper.
Q (marked blue) denotes the wire carrying current in a plane coming out of the paper.
Also from the figure, we can see that
 PA4 = QA4  A4A3P = A4A3Q = 90° A4PA3 =  A4QA3 = 45° PA4A3 =  QA4A3 = 45° PA4Q = 90°   
The magnetic field at A1 due to current in the wires is given by
B=μ0I2πd-μ0I2πd'    ...(1)

B= 2×10-7×102×10-2-2×10-7×106×10-2        = 1-13×10-4        = 0.67×10-4 T

Similarly, we get the magnetic field at A2 using eq. (1).
B= 2×10-7×101×10-2+2×10-7×103×10-2   =83×10-4 T   =2.67×10-4 T

Now,
Magnetic field at A3:
B = 2×10-7×102×10-2+2×10-7×102×10-2   = 2×10-4 T

Magnetic field at A4:
Separation of point A4from the wire on the left side, d = 22+22 =22 cm
Separation of point A4from the wire on the right side, d' =22+22 =22 cm
Thus, the magnetic field at A4 due to current in the wires is given by
B = 2×10-7×1022×10-22+2×10-7×1022×10-22   =1×10-4 T

Answer:

Given:
Magnitude of currents, I1= I2 = 10 A
Separation of the point from the wires, d = 2 cm
Thus, the magnetic field due to current in the wire is given by
B1=B2=μ0I2πd


In the figure, dotted circle shows the magnetic field lines due to current carrying wire placed in a plane perpendicular to the plane of the paper.
From the figure, we can see that PI1I2  is an equilateral triangle.
I1PI2 = 60°
Angle between the magnetic fields due to current in the wire, θ = 60°
∴ Required magnetic field at P
Bnet=B12+B22+2B1B2cosθ
   = 2×10-7×102×10-22+2×10-7×102×10-22+2×10-7×102×10-2+2×10-7×102×10-2cos60°
   = (10-4)+(10-4)2+2(10-4)(10-4)×12= 3×10-4 T= 1.732×10-4 T

Page No 250:

Question 11:

Given:
Magnitude of currents, I1= I2 = 10 A
Separation of the point from the wires, d = 2 cm
Thus, the magnetic field due to current in the wire is given by
B1=B2=μ0I2πd


In the figure, dotted circle shows the magnetic field lines due to current carrying wire placed in a plane perpendicular to the plane of the paper.
From the figure, we can see that PI1I2  is an equilateral triangle.
I1PI2 = 60°
Angle between the magnetic fields due to current in the wire, θ = 60°
∴ Required magnetic field at P
Bnet=B12+B22+2B1B2cosθ
   = 2×10-7×102×10-22+2×10-7×102×10-22+2×10-7×102×10-2+2×10-7×102×10-2cos60°
   = (10-4)+(10-4)2+2(10-4)(10-4)×12= 3×10-4 T= 1.732×10-4 T

Answer:

Given:
Magnitude of current, I = 5 A
Separation of the point from the wire, d = 1 m
Thus, the magnitude of magnetic field due to current in the wires is given by
B1=B2=μ0I2πd


(a) At point (1 m, 1 m), the magnetic fields due to the wires are the same in magnitude, but they are opposite in direction.
Hence, the net magnetic field is zero.

(b) At point (−1 m, 1 m), the magnetic fields due to the wires are in upward direction.
Bnet = B1+B2              = 2×10-7×51+2×10-7×51
               = 2 × 10−6 T      (Along the z-axis)

(c) At point (−1 m, −1 m), the magnetic fields due to the wires are the same in magnitude, but they are opposite in direction.
Hence, the net magnetic field is zero.

(d) At point (1 m, −1 m), the magnetic fields due to the wires are in upward direction.
Bnet = B1+B2              = 2×10-7×51+2×10-7×51
               = 2 × 10−6 T    (Along the negative z-axis)

Page No 250:

Question 12:

Given:
Magnitude of current, I = 5 A
Separation of the point from the wire, d = 1 m
Thus, the magnitude of magnetic field due to current in the wires is given by
B1=B2=μ0I2πd


(a) At point (1 m, 1 m), the magnetic fields due to the wires are the same in magnitude, but they are opposite in direction.
Hence, the net magnetic field is zero.

(b) At point (−1 m, 1 m), the magnetic fields due to the wires are in upward direction.
Bnet = B1+B2              = 2×10-7×51+2×10-7×51
               = 2 × 10−6 T      (Along the z-axis)

(c) At point (−1 m, −1 m), the magnetic fields due to the wires are the same in magnitude, but they are opposite in direction.
Hence, the net magnetic field is zero.

(d) At point (1 m, −1 m), the magnetic fields due to the wires are in upward direction.
Bnet = B1+B2              = 2×10-7×51+2×10-7×51
               = 2 × 10−6 T    (Along the negative z-axis)

Answer:


Given:

Let the horizontal wires placed at the bottom and top are denoted as W1 and W2 respectively.
Let the vertical wires placed at the right and left to point P are denoted as W3 and W4 respectively.
Magnitude of current, I = 5 A
(a) Consider point P.
Magnetic fields due to wires W1 and W2are the same in magnitude, but they are opposite in direction.
Magnetic fields due to wires W3 and W4are the same in magnitude, but they are opposite in direction.
Hence, the net magnetic field is zero.
Net magnetic field at P due to these four wires = 0

(b) Consider point Q1.
Due to wire W1, separation of point Q1from the wire (d) is 7.5 cm.
So, the magnetic field due to current in the wire is given by
BW1=μ0I2πd
       = 4 × 10−5T     (In upward direction)

Due to wire W2, separation of point Q1from the wire (d) is 2.5 cm.
So, the magnetic field due to current in the wire is given by
BW2=43×10-5 T    (In upward direction)

Due to wire W3, separation of point Q1from the wire (d) is 7.5 cm.
So, the magnetic field due to current in the wire is given by
BW3 = 4 × 10−5 T   (In upward direction)

Due to wire W4, separation of point Q1from the wire (d) is 2.5 cm.
So, the magnetic field due to current in the wire is given by
BW4=43×10-5 T   (In upward direction)

∴ Net magnetic field at point Q1
BQ1=4+43+4+43×10-5       =323×10-5      =1.06×10-4 T    (In upward direction)

At point Q2,
Magnetic field due to wire W1:
BW1= 4 × 10−5 T   (In upward direction)
Magnetic field due to wire W2:
BW2=43×10-5 T    (In upward direction)
Magnetic field due to wire W3:
BW3=43×10-5 T   (In downward direction)
Magnetic field due to wire W4:
BW4=4×10-5 T      (In downward direction)

∴ Net magnetic field at point Q2,
BQ2=0

Similarly, at point Q3, the magnetic field is 1.1 × 10−4T (in downward direction) and at point Q4, the magnetic field is zero.

Page No 250:

Question 13:


Given:

Let the horizontal wires placed at the bottom and top are denoted as W1 and W2 respectively.
Let the vertical wires placed at the right and left to point P are denoted as W3 and W4 respectively.
Magnitude of current, I = 5 A
(a) Consider point P.
Magnetic fields due to wires W1 and W2are the same in magnitude, but they are opposite in direction.
Magnetic fields due to wires W3 and W4are the same in magnitude, but they are opposite in direction.
Hence, the net magnetic field is zero.
Net magnetic field at P due to these four wires = 0

(b) Consider point Q1.
Due to wire W1, separation of point Q1from the wire (d) is 7.5 cm.
So, the magnetic field due to current in the wire is given by
BW1=μ0I2πd
       = 4 × 10−5T     (In upward direction)

Due to wire W2, separation of point Q1from the wire (d) is 2.5 cm.
So, the magnetic field due to current in the wire is given by
BW2=43×10-5 T    (In upward direction)

Due to wire W3, separation of point Q1from the wire (d) is 7.5 cm.
So, the magnetic field due to current in the wire is given by
BW3 = 4 × 10−5 T   (In upward direction)

Due to wire W4, separation of point Q1from the wire (d) is 2.5 cm.
So, the magnetic field due to current in the wire is given by
BW4=43×10-5 T   (In upward direction)

∴ Net magnetic field at point Q1
BQ1=4+43+4+43×10-5       =323×10-5      =1.06×10-4 T    (In upward direction)

At point Q2,
Magnetic field due to wire W1:
BW1= 4 × 10−5 T   (In upward direction)
Magnetic field due to wire W2:
BW2=43×10-5 T    (In upward direction)
Magnetic field due to wire W3:
BW3=43×10-5 T   (In downward direction)
Magnetic field due to wire W4:
BW4=4×10-5 T      (In downward direction)

∴ Net magnetic field at point Q2,
BQ2=0

Similarly, at point Q3, the magnetic field is 1.1 × 10−4T (in downward direction) and at point Q4, the magnetic field is zero.

Answer:

As shown in the figure, points P, Q, R and S lie on a circle of radius d.
Let the wires be named W1 and W2.


Now,
At point P, the magnetic field due to wire W1 is given by
B1 = 0
At point P, the magnetic field due to wire W2 is given by
B2=μ0i4πd  (Perpendicular to the plane in outward direction)
Bnet=μ0i4πd (Perpendicular to the plane in outward direction)

At point Q, the magnetic field due to wire W1 is given by
B1=μ0i4πd  (Perpendicular to the plane in inward direction)
At point Q, the magnetic field due to wire W2 is given by
B2 = 0
Bnet=μ0i4πd (Perpendicular to the plane in inward direction)

At point R, the magnetic field due to wire W1 is given by
B1 = 0
At point R, the magnetic field due to wire W2 is given by
B2=μ0i4πd  (Perpendicular to the plane in inward direction)
Bnet=μ0i4πd (Perpendicular to the plane in inward direction)

At point S, the magnetic field due to wire W1 is given by
B1=μ0i4πd  (Perpendicular to the plane in outward direction)
At point S, the magnetic field due to wire W2 is given by
B2 = 0
Bnet=μ0i4πd (Perpendicular to the plane in outward direction)
Hence, the magnitude of the magnetic field at points P, Q,  R and S is μ0i4πd.

Page No 250:

Question 14:

As shown in the figure, points P, Q, R and S lie on a circle of radius d.
Let the wires be named W1 and W2.


Now,
At point P, the magnetic field due to wire W1 is given by
B1 = 0
At point P, the magnetic field due to wire W2 is given by
B2=μ0i4πd  (Perpendicular to the plane in outward direction)
Bnet=μ0i4πd (Perpendicular to the plane in outward direction)

At point Q, the magnetic field due to wire W1 is given by
B1=μ0i4πd  (Perpendicular to the plane in inward direction)
At point Q, the magnetic field due to wire W2 is given by
B2 = 0
Bnet=μ0i4πd (Perpendicular to the plane in inward direction)

At point R, the magnetic field due to wire W1 is given by
B1 = 0
At point R, the magnetic field due to wire W2 is given by
B2=μ0i4πd  (Perpendicular to the plane in inward direction)
Bnet=μ0i4πd (Perpendicular to the plane in inward direction)

At point S, the magnetic field due to wire W1 is given by
B1=μ0i4πd  (Perpendicular to the plane in outward direction)
At point S, the magnetic field due to wire W2 is given by
B2 = 0
Bnet=μ0i4πd (Perpendicular to the plane in outward direction)
Hence, the magnitude of the magnetic field at points P, Q,  R and S is μ0i4πd.

Answer:

Let AB be the wire of length x with midpoint O.
Given:
Magnitude of current = i
Separation of the point from the wire = d

Now,
The magnetic field on a perpendicular bisector is given by
B = μ0i4πd(sinθ+sinθ)
 B = μ0i4πd2xx2+4d2
So, if  d > > x (neglecting x), then
B=μ0i4πd2x2dB1d2
And, if d < < x (neglecting d), then
B = μ0i4πd2xxB1d

Page No 250:

Question 15:

Let AB be the wire of length x with midpoint O.
Given:
Magnitude of current = i
Separation of the point from the wire = d

Now,
The magnetic field on a perpendicular bisector is given by
B = μ0i4πd(sinθ+sinθ)
 B = μ0i4πd2xx2+4d2
So, if  d > > x (neglecting x), then
B=μ0i4πd2x2dB1d2
And, if d < < x (neglecting d), then
B = μ0i4πd2xxB1d

Answer:


Let AB be the wire of length 10 cm and P be the required point.
Given:
Magnitude of current, i = 10 A
The angles made by points A and B with point P are θ1=30° and θ2=30°, respectively.
∴ Separation of the point from the wire, d32a=32×10=53 cm

Thus, the magnetic field due to current in the wire is given by
B = μ0i4πd(sinθ1+sinθ2)= 10-7×1053×10-212+12= 11.54×10-6 T

Page No 250:

Question 16:


Let AB be the wire of length 10 cm and P be the required point.
Given:
Magnitude of current, i = 10 A
The angles made by points A and B with point P are θ1=30° and θ2=30°, respectively.
∴ Separation of the point from the wire, d32a=32×10=53 cm

Thus, the magnetic field due to current in the wire is given by
B = μ0i4πd(sinθ1+sinθ2)= 10-7×1053×10-212+12= 11.54×10-6 T

Answer:

Given:
Magnitude of current = i
Separation of the point from the wire = d


Thus, the magnetic field due to current in the long wire is given by
B1=μ0i2πd
Also, the magnetic field due to a section of length l on a perpendicular bisector is given by
B2=μ0i4πd2ll2+4d2

μ0i l4πd2dl2d2+4Neglecting l2d2 very small, we getB2=μ0i l4πd2×22    =2μ0i l4πd2
Now,
B1 > B2
According to the question,
B1-B2B1=1100B2=0.99 B12μ0il4πd2=0.99×μ0i2πddl=1.4141.98=0.71

Page No 250:

Question 17:

Given:
Magnitude of current = i
Separation of the point from the wire = d


Thus, the magnetic field due to current in the long wire is given by
B1=μ0i2πd
Also, the magnetic field due to a section of length l on a perpendicular bisector is given by
B2=μ0i4πd2ll2+4d2

μ0i l4πd2dl2d2+4Neglecting l2d2 very small, we getB2=μ0i l4πd2×22    =2μ0i l4πd2
Now,
B1 > B2
According to the question,
B1-B2B1=1100B2=0.99 B12μ0il4πd2=0.99×μ0i2πddl=1.4141.98=0.71

Answer:

Let the currents in wires ABC and ADC be i1 and i2, respectively.
The resistances in wires ABC and ADC are r and 2r, respectively.

 i1i2 = 21 i1-2i2 = 0    ...1
And,
i1+i2 = i    ...2
Using (1) and (2), we get 
i1 = 2i3 and i2 = i3
The angles made by points A and D with point O are θ1=45° and θ2=45°, respectively.
Separation of the point from the wire, d = a/2
Now,
The magnetic field due to current in wire AD is given by
B=μ0i24πd(sinθ1+sinθ2)B=μ0i34πa2(sin45+sin45)     
The magnetic field at centre due to wire ADC is given by
B' = 2B = μ04πi3aa2×4×2    =2μ0i3πa
(Perpendicular to the plane in outward direction)

The magnetic field at centre due to wire ABC is given by
B'' = μ4π2i3aa2×4×2=22μ0i3πa
(Perpendicular to the plane in inward direction)
Bnet = B''-B' = 2μ0i3πa
(Perpendicular to the plane in inward direction)

Page No 250:

Question 18:

Let the currents in wires ABC and ADC be i1 and i2, respectively.
The resistances in wires ABC and ADC are r and 2r, respectively.

 i1i2 = 21 i1-2i2 = 0    ...1
And,
i1+i2 = i    ...2
Using (1) and (2), we get 
i1 = 2i3 and i2 = i3
The angles made by points A and D with point O are θ1=45° and θ2=45°, respectively.
Separation of the point from the wire, d = a/2
Now,
The magnetic field due to current in wire AD is given by
B=μ0i24πd(sinθ1+sinθ2)B=μ0i34πa2(sin45+sin45)     
The magnetic field at centre due to wire ADC is given by
B' = 2B = μ04πi3aa2×4×2    =2μ0i3πa
(Perpendicular to the plane in outward direction)

The magnetic field at centre due to wire ABC is given by
B'' = μ4π2i3aa2×4×2=22μ0i3πa
(Perpendicular to the plane in inward direction)
Bnet = B''-B' = 2μ0i3πa
(Perpendicular to the plane in inward direction)

Answer:

B at P due to AD = μ04π.i2.4d2.aa2a22+a42+a2a22+3a42along•
μ0i4πaa2a22+a42+a2a22+3a42along•

B at P due to AC=μ04π.i2.169a2.a.23a43a42+a22
=4μ0i9πa3a4a42+3a22along•
B at P due to AB = μ04π.i2.169a2.a.23a43a42+a22along•
B at P due to BC = μ04π.i2.4a2.a.a2a22+a42+a2a22+3a42 along•
=μ0i2πaa2a22+a42+a2a22+3a42along
So, net magnetic field at point P.
B=4μ0iπaa4a22+a42-4μ0i9πa3a4a22+3a42=4μ0iπa14114+116-4μ0i9πa.34114+916
=4μ0i4πa45-μ0i3πa413413 along•
=4μ0i2πa15-1313 along•
=2μ0iπa15-1313 along•



Page No 251:

Question 19:

B at P due to AD = μ04π.i2.4d2.aa2a22+a42+a2a22+3a42along•
μ0i4πaa2a22+a42+a2a22+3a42along•

B at P due to AC=μ04π.i2.169a2.a.23a43a42+a22
=4μ0i9πa3a4a42+3a22along•
B at P due to AB = μ04π.i2.169a2.a.23a43a42+a22along•
B at P due to BC = μ04π.i2.4a2.a.a2a22+a42+a2a22+3a42 along•
=μ0i2πaa2a22+a42+a2a22+3a42along
So, net magnetic field at point P.
B=4μ0iπaa4a22+a42-4μ0i9πa3a4a22+3a42=4μ0iπa14114+116-4μ0i9πa.34114+916
=4μ0i4πa45-μ0i3πa413413 along•
=4μ0i2πa15-1313 along•
=2μ0iπa15-1313 along•

Answer:

The loop ABCD can be considered as a circuit with two resistances in parallel, one along branch AB and other along branch ADC.

As, the sides of the loop are identical, their resistances are also same.

Let the resistance of each side be r.

The resistance of branch AB = r

The resistance of branch ADC = 3r

The current in the branches are calculated as:

iAB=i×3r3r+r=3i4

iADC=i×r3r+r=i4
As current follow the least resistive path so

Current in branch AB = 3i4
Current in branch ADC = i4
At the centre of the loop:

Magnetic field due to wire AD, DC and CB will be into the plane of paper according to right hand thumb rule.
Magnetic field due to wire AB will be out of the plane of paper according to right hand thumb rule.
Net magnetic field at the centre = BAD +BDC +BCB − BAB  which will be out of the plane of paper.
As, perpendicular distance of the centre from every wire will be equal to a2 and angle made by corner points of each side at the centre is 45°.
BAD = μ0(i4)4π(a2)(sin45°+sin45°)= μ0i8πaBAD = BDC=BCBBAD + BDC+BCB =B, = 3μ0i8πaBAB = μ0(3i4)4π(a2)(sin45°+sin45°)= 3μ0i8πaBnet =B, - BAB=0

Page No 251:

Question 20:

The loop ABCD can be considered as a circuit with two resistances in parallel, one along branch AB and other along branch ADC.

As, the sides of the loop are identical, their resistances are also same.

Let the resistance of each side be r.

The resistance of branch AB = r

The resistance of branch ADC = 3r

The current in the branches are calculated as:

iAB=i×3r3r+r=3i4

iADC=i×r3r+r=i4
As current follow the least resistive path so

Current in branch AB = 3i4
Current in branch ADC = i4
At the centre of the loop:

Magnetic field due to wire AD, DC and CB will be into the plane of paper according to right hand thumb rule.
Magnetic field due to wire AB will be out of the plane of paper according to right hand thumb rule.
Net magnetic field at the centre = BAD +BDC +BCB − BAB  which will be out of the plane of paper.
As, perpendicular distance of the centre from every wire will be equal to a2 and angle made by corner points of each side at the centre is 45°.
BAD = μ0(i4)4π(a2)(sin45°+sin45°)= μ0i8πaBAD = BDC=BCBBAD + BDC+BCB =B, = 3μ0i8πaBAB = μ0(3i4)4π(a2)(sin45°+sin45°)= 3μ0i8πaBnet =B, - BAB=0

Answer:

Let current 2I enter the circuit.
Since the wire is uniform, the current will be equally divided at point A (as shown in the figure).

Now,
Magnetic field at P due to wire AB = B (say)
(Perpendicular to the plane in outward direction)

Magnetic field at P due to wire BD = B' (say)
(Perpendicular to the plane in outward direction)

Magnetic field at P due to wire AC = Magnetic field at P due to wire AB = B
(Perpendicular to the plane in inward direction)

Magnetic field at P due to wire CD = Magnetic field at P due to wire BD = B'
(Perpendicular to the plane in inward direction)

∴ Net magnetic field at P = B + B' − B − B' = 0

Page No 251:

Question 21:

Let current 2I enter the circuit.
Since the wire is uniform, the current will be equally divided at point A (as shown in the figure).

Now,
Magnetic field at P due to wire AB = B (say)
(Perpendicular to the plane in outward direction)

Magnetic field at P due to wire BD = B' (say)
(Perpendicular to the plane in outward direction)

Magnetic field at P due to wire AC = Magnetic field at P due to wire AB = B
(Perpendicular to the plane in inward direction)

Magnetic field at P due to wire CD = Magnetic field at P due to wire BD = B'
(Perpendicular to the plane in inward direction)

∴ Net magnetic field at P = B + B' − B − B' = 0

Answer:

Let ABC be the equilateral triangle with side l/3 and centre M.
(a)
In AOB,AO=l32-I62     =l19-136=l4-136=l112MO=13×l112=l63
The angles made by points B and C with centre M are θ1=60° and θ2=60°, respectively.
Separation of the point from the wire, d = MO = l63
Thus, the magnetic field due to current in wire BC is given by
B = μ0i4πd(sin θ1+sin θ2)B=μ0i4πl63(sin 60+sin 60)     
 B=μ0i4πl63×3
Now,
Net magnetic field at M = Magnetic field due to wire BC + Magnetic field due to wire CA + Magnetic field due to wire AB
Since all wires are the same,
Bnet = 3B=27μ0iπl
It is perpendicular to the plane in outward direction if the current is anticlockwise and perpendicular to the plane in inward direction if the current is clockwise.

(b)
The angles made by points B and C with centre M are θ1=45° and θ2=45°, respectively.
Separation of the point from the wire, d = l/8
Thus, the magnetic field due to current in wire BC is given by
B=μ0i4πd(sin θ1+sin θ2)B=μ0i4πl8(sin 45+sin 45)=22μ0iπl     
Since all wires are the same,
Net magnetic field at M = 4 × Magnetic field due to wire BC
Bnet = 4B =82μ0iπl

Page No 251:

Question 22:

Let ABC be the equilateral triangle with side l/3 and centre M.
(a)
In AOB,AO=l32-I62     =l19-136=l4-136=l112MO=13×l112=l63
The angles made by points B and C with centre M are θ1=60° and θ2=60°, respectively.
Separation of the point from the wire, d = MO = l63
Thus, the magnetic field due to current in wire BC is given by
B = μ0i4πd(sin θ1+sin θ2)B=μ0i4πl63(sin 60+sin 60)     
 B=μ0i4πl63×3
Now,
Net magnetic field at M = Magnetic field due to wire BC + Magnetic field due to wire CA + Magnetic field due to wire AB
Since all wires are the same,
Bnet = 3B=27μ0iπl
It is perpendicular to the plane in outward direction if the current is anticlockwise and perpendicular to the plane in inward direction if the current is clockwise.

(b)
The angles made by points B and C with centre M are θ1=45° and θ2=45°, respectively.
Separation of the point from the wire, d = l/8
Thus, the magnetic field due to current in wire BC is given by
B=μ0i4πd(sin θ1+sin θ2)B=μ0i4πl8(sin 45+sin 45)=22μ0iπl     
Since all wires are the same,
Net magnetic field at M = 4 × Magnetic field due to wire BC
Bnet = 4B =82μ0iπl

Answer:

Let CAB be the wire making an angle α, P be the point on the bisector of this angle situated at a distance x from the vertex A and d be the perpendicular distance of AC and AB from P.

From the figure,
sin α2=dx
d = xsin α2
The angles made by points A and C with point P are θ1 = 90-α2 and θ2=90°, respectively.
Separation of the point from the wire, d = xsin α2
Thus, the magnetic field due to current in wire AC is given by
B = μ0i4πd(sin θ1+sin θ2)=μ0i4πxsin α2sin 90-α2+sin 90       
μ0i4πxsin α2cos α2+1=μ0i2cos2 α44πx2sin α4cos α4 = μ0icot α44πx
Now, the magnetic field due to wires AC and AB is given by
Bnet=2B=μ0icot α42πx

Page No 251:

Question 23:

Let CAB be the wire making an angle α, P be the point on the bisector of this angle situated at a distance x from the vertex A and d be the perpendicular distance of AC and AB from P.

From the figure,
sin α2=dx
d = xsin α2
The angles made by points A and C with point P are θ1 = 90-α2 and θ2=90°, respectively.
Separation of the point from the wire, d = xsin α2
Thus, the magnetic field due to current in wire AC is given by
B = μ0i4πd(sin θ1+sin θ2)=μ0i4πxsin α2sin 90-α2+sin 90       
μ0i4πxsin α2cos α2+1=μ0i2cos2 α44πx2sin α4cos α4 = μ0icot α44πx
Now, the magnetic field due to wires AC and AB is given by
Bnet=2B=μ0icot α42πx

Answer:

Let the angles made by points A and B with point O be θ1 and θ2, respectively.
And,
θ1 = θ2 = θ    
Separation of the point from the wire, d = b/2


In AOM,AO = b22+l22=12b2+l2
And,
sin θ=l2121b2+l2=lb2+l2
Thus, the magnetic field due to current in wire AB is given by
B=μ0i4πd(sin θ1+sin θ2)B=μ0i4πd×2sin θ      
B =μ0i4πb2×2ll2+b2        =μ0iπbll2+b2
Similarly, the magnetic field due to current in wire BC is given by
B' = μ0iπlbl2+b2
Now,
Magnetic field due to current in wire CD = Magnetic field due to current in wire AB = B
And,
Magnetic field due to current in wire DA = Magnetic field due to current in wire BC = B'

Bnet=2B+B'= 2μ0iπbll2+b2+μ0iπlbl2+b2=2μ0iπl2+b2lb+bl=2μ0ib2+l2πbl

Page No 251:

Question 24:

Let the angles made by points A and B with point O be θ1 and θ2, respectively.
And,
θ1 = θ2 = θ    
Separation of the point from the wire, d = b/2


In AOM,AO = b22+l22=12b2+l2
And,
sin θ=l2121b2+l2=lb2+l2
Thus, the magnetic field due to current in wire AB is given by
B=μ0i4πd(sin θ1+sin θ2)B=μ0i4πd×2sin θ      
B =μ0i4πb2×2ll2+b2        =μ0iπbll2+b2
Similarly, the magnetic field due to current in wire BC is given by
B' = μ0iπlbl2+b2
Now,
Magnetic field due to current in wire CD = Magnetic field due to current in wire AB = B
And,
Magnetic field due to current in wire DA = Magnetic field due to current in wire BC = B'

Bnet=2B+B'= 2μ0iπbll2+b2+μ0iπlbl2+b2=2μ0iπl2+b2lb+bl=2μ0ib2+l2πbl

Answer:



(a)
Using figure,
For a regure polygon of n-sides, the angle subtended at the centre is 2πntanθ=l2xx=l2tanθconsidering angle to be small l2=πrnUsing Biot-Savart law for one sideB=μ04πidlsinθx2B=μ04πi(sinθ+sinθ)x2=μ0i2tanθ2sinθ4πl   Putting value of rB=μ0i2ntanπn2sinπn4π2πr       Putting value of lFor n-sided polygonB'=nBB'=μ0in2tanπnsinπn2π2r
(b)
when n→ ∞, polygon becomes a circle with radius r
and magnetic field will become
B=μ0i2r

Page No 251:

Question 25:



(a)
Using figure,
For a regure polygon of n-sides, the angle subtended at the centre is 2πntanθ=l2xx=l2tanθconsidering angle to be small l2=πrnUsing Biot-Savart law for one sideB=μ04πidlsinθx2B=μ04πi(sinθ+sinθ)x2=μ0i2tanθ2sinθ4πl   Putting value of rB=μ0i2ntanπn2sinπn4π2πr       Putting value of lFor n-sided polygonB'=nBB'=μ0in2tanπnsinπn2π2r
(b)
when n→ ∞, polygon becomes a circle with radius r
and magnetic field will become
B=μ0i2r

Answer:




By appluing Kirchoff voltagee law,we can see that the current in the circuit is zero.
Net current in the circuit = 0
As
magnetic field is always proportional to the current flowing in the circuit hence,
Net magnetic field at point P = 0
Field at point P is independent of the values of the resistances in the circuit.

Page No 251:

Question 26:




By appluing Kirchoff voltagee law,we can see that the current in the circuit is zero.
Net current in the circuit = 0
As
magnetic field is always proportional to the current flowing in the circuit hence,
Net magnetic field at point P = 0
Field at point P is independent of the values of the resistances in the circuit.

Answer:

Given:
Magnitude of current in both wires, i1 = i2 = 20 A
Force acting on 0.1 m of the second wire, F = 2.0 × 10−5 N
∴ Force per unit length = 2×10-50.1 = 2.0 × 10−4 N/m
Now,
Let the separation between the two wires be d.
Thus, the force per unit length is given by
 Fl=μ0i1i22πd2.0×10-4=2×10-7×20×20dd=0.4 m=40 cm

Page No 251:

Question 27:

Given:
Magnitude of current in both wires, i1 = i2 = 20 A
Force acting on 0.1 m of the second wire, F = 2.0 × 10−5 N
∴ Force per unit length = 2×10-50.1 = 2.0 × 10−4 N/m
Now,
Let the separation between the two wires be d.
Thus, the force per unit length is given by
 Fl=μ0i1i22πd2.0×10-4=2×10-7×20×20dd=0.4 m=40 cm

Answer:

Let wires W1, W2 and W3 be arranged as shown in the figure.

Given:
Magnitude of current in each wire, i1= i2 = i3 = 10 A
The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by
 Fl=μ0i1i22πd
So, for wire W1,
Fl =Flby wire W2+Flby wire W3 = μ0×10×102π×5×10-2+μ0×10×102π×10×10-2= 2×10-7×1025×10-2+2×10-7×10210×10-2= 4×10-4+2×10-4= 6×10-4 N 

For wire W2,
Fl=Fl by wire W1-Fl by wire W3 = μ0×10×102π×5×10-2-μ0×10×102π×5×10-2 = 0

For wire W3,
Fl = Fl by wire W1+Fl by wire W2 =μ0×10×102π×5×10-2+μ0×10×102π×10×10-2
 = 6 × 10−4 N

Page No 251:

Question 28:

Let wires W1, W2 and W3 be arranged as shown in the figure.

Given:
Magnitude of current in each wire, i1= i2 = i3 = 10 A
The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by
 Fl=μ0i1i22πd
So, for wire W1,
Fl =Flby wire W2+Flby wire W3 = μ0×10×102π×5×10-2+μ0×10×102π×10×10-2= 2×10-7×1025×10-2+2×10-7×10210×10-2= 4×10-4+2×10-4= 6×10-4 N 

For wire W2,
Fl=Fl by wire W1-Fl by wire W3 = μ0×10×102π×5×10-2-μ0×10×102π×5×10-2 = 0

For wire W3,
Fl = Fl by wire W1+Fl by wire W2 =μ0×10×102π×5×10-2+μ0×10×102π×10×10-2
 = 6 × 10−4 N

Answer:

Let the third wire W3 having current i in upward direction be placed x cm from the 10 A current wire.


The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by
 Fl = μ0i1i22πd
According to the question, wire W3experiences no magnetic force.
∴ Fl due to wire W1=Fl due to wire W2 
μ010i2πx=μ040i2π(10-x)10-x=4xx=2 cm
Thus, wire W3 is placed 2 cm from the 10 A current wire.

Page No 251:

Question 29:

Let the third wire W3 having current i in upward direction be placed x cm from the 10 A current wire.


The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by
 Fl = μ0i1i22πd
According to the question, wire W3experiences no magnetic force.
∴ Fl due to wire W1=Fl due to wire W2 
μ010i2πx=μ040i2π(10-x)10-x=4xx=2 cm
Thus, wire W3 is placed 2 cm from the 10 A current wire.

Answer:

Since wires AB, CD and EF have identical resistance, the current (30 A) gets equally distributed in them, that is, 10 A in each wire.

The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by
 Fl = μ0i1i22πd
∴ Magnetic force per unit length of AB = Force due to current in CD + Force due to current in EF
Fl =  μ0×10×102π×1×10-2+μ0×10×102π×2×10-2=  2×10-7×10210-2+2×10-7×1022×10-2= 2×10-3+10-3= 3×10-3 N/m 
Similarly,
Magnetic force per unit length of CD = Force due to current in AB − Force due to current in EF  
∵ Force on CD due to current in AB = Force due to current in EF  
∴ Magnetic force per unit length of CD = 0

Page No 251:

Question 30:

Since wires AB, CD and EF have identical resistance, the current (30 A) gets equally distributed in them, that is, 10 A in each wire.

The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by
 Fl = μ0i1i22πd
∴ Magnetic force per unit length of AB = Force due to current in CD + Force due to current in EF
Fl =  μ0×10×102π×1×10-2+μ0×10×102π×2×10-2=  2×10-7×10210-2+2×10-7×1022×10-2= 2×10-3+10-3= 3×10-3 N/m 
Similarly,
Magnetic force per unit length of CD = Force due to current in AB − Force due to current in EF  
∵ Force on CD due to current in AB = Force due to current in EF  
∴ Magnetic force per unit length of CD = 0

Answer:

Given:
Magnitude of current, i1 = 10 A
Separation between two wires, d = 5 mm
Linear mass density of the second wire, λ = 1.0 × 10−4 kgm−1
Now,
Let i2be the current in the second wire in opposite direction.
Thus, the magnetic force per unit length on the wire due to a parallel current-carrying wire is given by
 Fml = μ0i1i22πd  (upwards)
Also,
Weight of the second wire, W = mg
Weight per unit length of the second wire, Wl = λg (downwards)

Now, according to the question,
Fml = Wlμ0i1i22πd=λg
2×10-7×50×i25×10-3 = 1×10-4×9.8i2=9.8×10-720×10-7 = 0.49 A

Page No 251:

Question 31:

Given:
Magnitude of current, i1 = 10 A
Separation between two wires, d = 5 mm
Linear mass density of the second wire, λ = 1.0 × 10−4 kgm−1
Now,
Let i2be the current in the second wire in opposite direction.
Thus, the magnetic force per unit length on the wire due to a parallel current-carrying wire is given by
 Fml = μ0i1i22πd  (upwards)
Also,
Weight of the second wire, W = mg
Weight per unit length of the second wire, Wl = λg (downwards)

Now, according to the question,
Fml = Wlμ0i1i22πd=λg
2×10-7×50×i25×10-3 = 1×10-4×9.8i2=9.8×10-720×10-7 = 0.49 A

Answer:

Given:
Current in the loop, i1 = 6 A
Current in the wire, i2 = 10 A

Now, consider an element on PQ of width dx at a distance x from the wire.


Force on the element is given by
dF=μ0i1i22πxdx
Force acting on part PQ is given by
FPQ=μ0i1i22π13dxx       =2×10-7×6×10[lnx]13       =120×10-7ln3 NSimilarly, FRS=μ0i1i22π31dxx       =120×10-7ln 13       =-120×10-7ln3 N
Both forces are equal in magnitude, but they are opposite in direction.

(b) The magnetic field intensity due to wire on SP is given by
B=μ0i22πr

Force on part SP is given by
FSP=i1Bl       =μ0i1i22πlr      =μ0i1i22π21
(Towards right)

Force on part RQ is given by
FRQ=i1Bl       =μ0i1i22πlr      =μ0i1i22π23
(Towards left)

Thus, the net force on the loop is given by
Fnet=FSP-FRQ       =μ0i1i22π21-23      =2×10-7×6×10×43      =16×10-6 N
(Towards right)

Page No 251:

Question 32:

Given:
Current in the loop, i1 = 6 A
Current in the wire, i2 = 10 A

Now, consider an element on PQ of width dx at a distance x from the wire.


Force on the element is given by
dF=μ0i1i22πxdx
Force acting on part PQ is given by
FPQ=μ0i1i22π13dxx       =2×10-7×6×10[lnx]13       =120×10-7ln3 NSimilarly, FRS=μ0i1i22π31dxx       =120×10-7ln 13       =-120×10-7ln3 N
Both forces are equal in magnitude, but they are opposite in direction.

(b) The magnetic field intensity due to wire on SP is given by
B=μ0i22πr

Force on part SP is given by
FSP=i1Bl       =μ0i1i22πlr      =μ0i1i22π21
(Towards right)

Force on part RQ is given by
FRQ=i1Bl       =μ0i1i22πlr      =μ0i1i22π23
(Towards left)

Thus, the net force on the loop is given by
Fnet=FSP-FRQ       =μ0i1i22π21-23      =2×10-7×6×10×43      =16×10-6 N
(Towards right)

Answer:

Given:
No. of turns, n = 1
Magnitude of current, i = 5.00 A
Now, let the radius of the loop be r.
Thus, the magnetic field at the centre due to the current in the loop is given by
B = μ0i2r
0.2×10-3 = 4π×10-7×52rr=1.57×10-2 m=1.57 cm

Page No 251:

Question 33:

Given:
No. of turns, n = 1
Magnitude of current, i = 5.00 A
Now, let the radius of the loop be r.
Thus, the magnetic field at the centre due to the current in the loop is given by
B = μ0i2r
0.2×10-3 = 4π×10-7×52rr=1.57×10-2 m=1.57 cm

Answer:

Given:
No. of turns, n = 100
Radius of the loop, r = 5 cm = 0.05 m
Magnetic field intensity, B = 6.0 × 10−5 T
Now, let the magnitude of current be i.
Using B=μ0ni2r, we get6.0×10-7=4π×10-7×102×i2×0.05i=60×10-74π×10-7×102      =4.777×10-248 mA

Page No 251:

Question 34:

Given:
No. of turns, n = 100
Radius of the loop, r = 5 cm = 0.05 m
Magnetic field intensity, B = 6.0 × 10−5 T
Now, let the magnitude of current be i.
Using B=μ0ni2r, we get6.0×10-7=4π×10-7×102×i2×0.05i=60×10-74π×10-7×102      =4.777×10-248 mA

Answer:

Given:
Frequency of the electron = 3 × 105
Time taken by the electron to complete one revolution, T = 1Frequency
Current in the circle, i = qt
Radius of the loop, r = 0.5 A° = 0.5×10-10 m
Thus, the magnetic field at the centre due to the current in the loop is given by
B = μ0I2r
   = 4π×10-7×qT2×0.5×10-10    =4π×10-7×1.6×10-192×0.5×10-10×3×105   = 6×10-10 T

Page No 251:

Question 35:

Given:
Frequency of the electron = 3 × 105
Time taken by the electron to complete one revolution, T = 1Frequency
Current in the circle, i = qt
Radius of the loop, r = 0.5 A° = 0.5×10-10 m
Thus, the magnetic field at the centre due to the current in the loop is given by
B = μ0I2r
   = 4π×10-7×qT2×0.5×10-10    =4π×10-7×1.6×10-192×0.5×10-10×3×105   = 6×10-10 T

Answer:

As the centre of the loop, that is, point O, lies on the same line of two long, straight wires, the magnetic field at O due to each straight wire is zero.


Since wires ABC and ADC are identical, the current gets equally distributed in two parts at point A. So, the magnetic field due to ABC and ADC at O are equal in magnitude but are opposite in directions. (as shown in the figure).
∴ Net magnetic field at O = 0



Page No 252:

Question 36:

As the centre of the loop, that is, point O, lies on the same line of two long, straight wires, the magnetic field at O due to each straight wire is zero.


Since wires ABC and ADC are identical, the current gets equally distributed in two parts at point A. So, the magnetic field due to ABC and ADC at O are equal in magnitude but are opposite in directions. (as shown in the figure).
∴ Net magnetic field at O = 0

Answer:

No. of turns: n1 = 50 and n2 = 100
Magnitude of currents: i1 = i2 = 2 A
Radii of loops: r1 = 5 cm and r2 = 10 cm
(a) In the same sense:

The magnetic field intensity at the centre is given by
B = μ0n1i12r1+μ0n2i22r2= 4π×10-7×50×22×5×10-2+4π×10-7×100×22×10×10-2= 4π×10-4+4π×10-4= 2×4π×10-6= 8π×10-3 T

(b) In the opposite sense:
The magnetic field intensity at the centre is given by

B = μ0n1i12r1-μ0n2i22r2    =4π×10-7×50×22×5×10-2-4π×10-7×100×22×10×10-2             =0

Page No 252:

Question 37:

No. of turns: n1 = 50 and n2 = 100
Magnitude of currents: i1 = i2 = 2 A
Radii of loops: r1 = 5 cm and r2 = 10 cm
(a) In the same sense:

The magnetic field intensity at the centre is given by
B = μ0n1i12r1+μ0n2i22r2= 4π×10-7×50×22×5×10-2+4π×10-7×100×22×10×10-2= 4π×10-4+4π×10-4= 2×4π×10-6= 8π×10-3 T

(b) In the opposite sense:
The magnetic field intensity at the centre is given by

B = μ0n1i12r1-μ0n2i22r2    =4π×10-7×50×22×5×10-2-4π×10-7×100×22×10×10-2             =0

Answer:

Given:
No. of turns: n1 = 50 and n2 = 100
Magnitude of currents: i1 = i2 = 2 A
Radii of loops: r1 = 5 cm and r2 = 10 cm
(a) In the same sense:

The magnetic field intensity at the centre due to C1 is given by
B1=μ0n1i12r1= 4π×10-7×50×22×5×10-2= 4π×10-4 T
(In the plane of paper in upward direction)

The magnetic field intensity at the centre due to C2 is given by
B2=μ0n2i22r2= 4π×10-7×100×22×10×10-2=4π×10-4 T

(In the plane of paper in upward direction)

In this case, magnetic fields due to C1 and C2 at the centre are along the same direction.
Thus, the net magnetic field is given by
Bnet = B1+B2= (4π×10-4)+(4π×10-4)= 8π×10-4 T= 25.12 mT
(b) When the direction of current in the two coils is opposite to each other then the magnetic fields will also point in opposite directions as shown in the figure. Hence, the net magnetic field will be obtained by the subtraction of the two magnetic fields.

Bnet = B1-B2= (4π×10-4)-(4π×10-4)=0

Page No 252:

Question 38:

Given:
No. of turns: n1 = 50 and n2 = 100
Magnitude of currents: i1 = i2 = 2 A
Radii of loops: r1 = 5 cm and r2 = 10 cm
(a) In the same sense:

The magnetic field intensity at the centre due to C1 is given by
B1=μ0n1i12r1= 4π×10-7×50×22×5×10-2= 4π×10-4 T
(In the plane of paper in upward direction)

The magnetic field intensity at the centre due to C2 is given by
B2=μ0n2i22r2= 4π×10-7×100×22×10×10-2=4π×10-4 T

(In the plane of paper in upward direction)

In this case, magnetic fields due to C1 and C2 at the centre are along the same direction.
Thus, the net magnetic field is given by
Bnet = B1+B2= (4π×10-4)+(4π×10-4)= 8π×10-4 T= 25.12 mT
(b) When the direction of current in the two coils is opposite to each other then the magnetic fields will also point in opposite directions as shown in the figure. Hence, the net magnetic field will be obtained by the subtraction of the two magnetic fields.

Bnet = B1-B2= (4π×10-4)-(4π×10-4)=0

Answer:

Given:
Magnitude of current in the loop, I = 10 A
Radius of the loop, r = 20 cm = 20 × 10−2 m
Thus, the magnetic field intensity at the centre is given by
B = μ0I2r
Now,
Velocity of the electron, v = 2 × 106 m/s
Angle between the velocity and the magnetic field intensity, θ = 30°
Thus, the magnetic force on the electron is given by
F = evBsin θ=1.6 ×10-19×2×106×μ0i2Rsin 30°= 1.6×10-19×2×106×4π×10-7×102×20×10-2×12=16π×10-19 N

Page No 252:

Question 39:

Given:
Magnitude of current in the loop, I = 10 A
Radius of the loop, r = 20 cm = 20 × 10−2 m
Thus, the magnetic field intensity at the centre is given by
B = μ0I2r
Now,
Velocity of the electron, v = 2 × 106 m/s
Angle between the velocity and the magnetic field intensity, θ = 30°
Thus, the magnetic force on the electron is given by
F = evBsin θ=1.6 ×10-19×2×106×μ0i2Rsin 30°= 1.6×10-19×2×106×4π×10-7×102×20×10-2×12=16π×10-19 N

Answer:

Given:
For the outer loop,
Magnitude of current = I
Radius of the loop = R
Thus, the magnetic field at the centre due to the larger loop is given by
B=μ0I2R
Let A be the area of the smaller loop and let current i pass through it.
Now,
Angle between the area vector of the smaller loop and the magnetic field due to the larger loop = 90°
Thus, the required torque is given by
Γ = i(A × B )
   = iABsin 90°
  =iπr2μ0I2R=μ0πr2Ii2R

Page No 252:

Question 40:

Given:
For the outer loop,
Magnitude of current = I
Radius of the loop = R
Thus, the magnetic field at the centre due to the larger loop is given by
B=μ0I2R
Let A be the area of the smaller loop and let current i pass through it.
Now,
Angle between the area vector of the smaller loop and the magnetic field due to the larger loop = 90°
Thus, the required torque is given by
Γ = i(A × B )
   = iABsin 90°
  =iπr2μ0I2R=μ0πr2Ii2R

Answer:

Given:
For the outer loop,
Magnitude of current = I
Radius of the loop = R
Thus, the magnetic field at the centre due to the larger loop is given by
B = μ0I2R
Let A be the area of the smaller loop and let current i pass through it.
Now,
Angle between the area vector of the smaller loop and the magnetic field due to the larger loop = 30°
Thus, the torque on the smaller loop is given by
Γ = i(A × B )
    = iABsin 30°
   =iπr2μ0I4R=μ0πr2Ii4R
If the smaller loop is held fixed in its position, then
Torque due to the magnetic field = Torque due to the external force at its periphery
Fr = μ0πr2Ii4RF = μ0πIir4R
This is the minimum magnitude of force to balance the given condition.

Page No 252:

Question 41:

Given:
For the outer loop,
Magnitude of current = I
Radius of the loop = R
Thus, the magnetic field at the centre due to the larger loop is given by
B = μ0I2R
Let A be the area of the smaller loop and let current i pass through it.
Now,
Angle between the area vector of the smaller loop and the magnetic field due to the larger loop = 30°
Thus, the torque on the smaller loop is given by
Γ = i(A × B )
    = iABsin 30°
   =iπr2μ0I4R=μ0πr2Ii4R
If the smaller loop is held fixed in its position, then
Torque due to the magnetic field = Torque due to the external force at its periphery
Fr = μ0πr2Ii4RF = μ0πIir4R
This is the minimum magnitude of force to balance the given condition.

Answer:

Given:
Magnitude of current, I = 5 A
Radius of the semi-circular wire, r = 10 cm
∴ Required magnetic field at the centre of curvature
B = 12×μ0i2r   = 10-7×510×10-2   = 5π×10-6   = 1.6×10-5 T

Page No 252:

Question 42:

Given:
Magnitude of current, I = 5 A
Radius of the semi-circular wire, r = 10 cm
∴ Required magnetic field at the centre of curvature
B = 12×μ0i2r   = 10-7×510×10-2   = 5π×10-6   = 1.6×10-5 T

Answer:



Given:
Magnitude of current, I = 6 A
Radius of the semi-circular wire, r = 10 cm
Angle subtended at the centre, θ = 120° = 2π3
∴ Required magnetic field at the centre of curvature
B= μ0i2rθ2π= 4×10-7×52×10×10-2×2π3×2π= 4π×10-6= 1.26×10-5 T

Page No 252:

Question 43:



Given:
Magnitude of current, I = 6 A
Radius of the semi-circular wire, r = 10 cm
Angle subtended at the centre, θ = 120° = 2π3
∴ Required magnetic field at the centre of curvature
B= μ0i2rθ2π= 4×10-7×52×10×10-2×2π3×2π= 4π×10-6= 1.26×10-5 T

Answer:

Given:
Magnitude of current = i
Radius of the loop = r
Magnetic field due to the loop at its centre, B l = μ0i2r
Let a straight wire carrying 4i current be placed at a distance x from the centre such that the magnetic fields of the loop and the wire are of equal magnitude but in opposite direction at O.

Magnetic field due to the wire at the centre of the loop, Bw = μ04i2πx
According to the question,
Bl = Bwμ0i2r = μ04i2πx x = 8r2π = 4rπ
This means that the wire is placed 4rπ from the centre of the loop (as shown in the figure).

Page No 252:

Question 44:

Given:
Magnitude of current = i
Radius of the loop = r
Magnetic field due to the loop at its centre, B l = μ0i2r
Let a straight wire carrying 4i current be placed at a distance x from the centre such that the magnetic fields of the loop and the wire are of equal magnitude but in opposite direction at O.

Magnetic field due to the wire at the centre of the loop, Bw = μ04i2πx
According to the question,
Bl = Bwμ0i2r = μ04i2πx x = 8r2π = 4rπ
This means that the wire is placed 4rπ from the centre of the loop (as shown in the figure).

Answer:

Number of turns, n = 200
Radius of the coil, r = 10 cm
Current in the coil, i = 2A
(a) Let the magnetic field at the centre of the coil is B.
As the relation for magnetic field at the centre of a circular coil is given by
B = 0i2r=200×4π×10-7×22×10×10-2= 2.51×10-3= 12.56 mT
(b) As magnetic field at any point P (say) on the axis of the circular coil is given by
BP = nμ0ir22(x2+r2)32
Where x is the distance of the point from the centre of the coil.
As per the question
12Bcentre = BP12nμ0i2r = nμ0ir22(x2+r2)32(x2+r2)32 = 2r3(x2+r2) = 41/3r2x2 + r2 = 1.58r2x =0.766 rx = ±7.66 cm
Magnetic field will drop to half of its value at the centre if the distance of that point from the centre of the coil along the axis of coil is equal to 7.66 cm.

Page No 252:

Question 45:

Number of turns, n = 200
Radius of the coil, r = 10 cm
Current in the coil, i = 2A
(a) Let the magnetic field at the centre of the coil is B.
As the relation for magnetic field at the centre of a circular coil is given by
B = 0i2r=200×4π×10-7×22×10×10-2= 2.51×10-3= 12.56 mT
(b) As magnetic field at any point P (say) on the axis of the circular coil is given by
BP = nμ0ir22(x2+r2)32
Where x is the distance of the point from the centre of the coil.
As per the question
12Bcentre = BP12nμ0i2r = nμ0ir22(x2+r2)32(x2+r2)32 = 2r3(x2+r2) = 41/3r2x2 + r2 = 1.58r2x =0.766 rx = ±7.66 cm
Magnetic field will drop to half of its value at the centre if the distance of that point from the centre of the coil along the axis of coil is equal to 7.66 cm.

Answer:

Given:
Magnitude of current, I = 5.0 A
Radius of the loop, r = 4.0 cm

    

(a) The magnetic field intensity B on point O at a distance x on the axial line is given by
B = μ02ir2(x2+r2)3/2
  =4π×10-7×5×16×10-42[(9×16)×10-4]3/2=4π×80×10-112×125×10-6=4.019×10-5 T    (in downward direction)

(b) The magnetic field intensity B on point O' at a distance x on the axial line is given by
B = μ02ir2(x2+r2)3/2
  =4π×10-7×5×16×10-42[(9×16)×10-4]3/2=4π×80×10-112×125×10-6=4.019×10-5 T        (in downward direction)

Page No 252:

Question 46:

Given:
Magnitude of current, I = 5.0 A
Radius of the loop, r = 4.0 cm

    

(a) The magnetic field intensity B on point O at a distance x on the axial line is given by
B = μ02ir2(x2+r2)3/2
  =4π×10-7×5×16×10-42[(9×16)×10-4]3/2=4π×80×10-112×125×10-6=4.019×10-5 T    (in downward direction)

(b) The magnetic field intensity B on point O' at a distance x on the axial line is given by
B = μ02ir2(x2+r2)3/2
  =4π×10-7×5×16×10-42[(9×16)×10-4]3/2=4π×80×10-112×125×10-6=4.019×10-5 T        (in downward direction)

Answer:




Given:
Magnitude of charges, q =  3.14 × 10−6 C
Radius of the ring, r=20 cm=20×10-2 m
Angular velocity of the ring, ω=60 rad/s
Time for 1 revolution = 2π60
Current,  i = qt = 3.14×10-6×602π=30×10-6 A
In the figure, E1 and E2 denotes the electric field at a point on the axis at a distance of 5.00 cm from the centre due to small element 1 and 2 of the ring respectively.
E is the resultant electric field due to the entire ring at a point on the axis at a distance of 5.00 cm from the centre.
The electric field at a point on the axis at a distance x from the centre is given by
E = xq4πε0(x2+r2)32

The magnetic field at a point on the axis at a distance x from the centre is given by
B = μ02ir2(x2+r2)3/2EB = xq4πε0(x2+r2)32μ02ir2(x2+r2)3/2= 9×109×3.14×10-6×2×(20.6)3×10-625×10-4×4π×10-14×12= 9×3.14×2×(20.6)325×4π×12= 1.88×1015 m/s

Page No 252:

Question 47:




Given:
Magnitude of charges, q =  3.14 × 10−6 C
Radius of the ring, r=20 cm=20×10-2 m
Angular velocity of the ring, ω=60 rad/s
Time for 1 revolution = 2π60
Current,  i = qt = 3.14×10-6×602π=30×10-6 A
In the figure, E1 and E2 denotes the electric field at a point on the axis at a distance of 5.00 cm from the centre due to small element 1 and 2 of the ring respectively.
E is the resultant electric field due to the entire ring at a point on the axis at a distance of 5.00 cm from the centre.
The electric field at a point on the axis at a distance x from the centre is given by
E = xq4πε0(x2+r2)32

The magnetic field at a point on the axis at a distance x from the centre is given by
B = μ02ir2(x2+r2)3/2EB = xq4πε0(x2+r2)32μ02ir2(x2+r2)3/2= 9×109×3.14×10-6×2×(20.6)3×10-625×10-4×4π×10-14×12= 9×3.14×2×(20.6)325×4π×12= 1.88×1015 m/s

Answer:

(a) The magnetic field inside any conducting tube is always zero.
∴ Magnetic field inside the tube at a distance r/2 from the surface = 0

(b) Let the point outside the tube with distance r2 be P.

∴ Net distance from centre, r' = r+r2=3r2
Consider an Amperian loop, as shown in the figure.
Length of the loop, l = 2π×32r=3πr
Current enclosed in the loop = i
On applying Ampere's law, we get
B.dl=μ0iB×3πr=μ0iB=μ0i3πr

Page No 252:

Question 48:

(a) The magnetic field inside any conducting tube is always zero.
∴ Magnetic field inside the tube at a distance r/2 from the surface = 0

(b) Let the point outside the tube with distance r2 be P.

∴ Net distance from centre, r' = r+r2=3r2
Consider an Amperian loop, as shown in the figure.
Length of the loop, l = 2π×32r=3πr
Current enclosed in the loop = i
On applying Ampere's law, we get
B.dl=μ0iB×3πr=μ0iB=μ0i3πr

Answer:




a) The magnetic field inside any conducting tube is always zero.
∴ Magnetic field just inside the tube is zero.

(b) Let the point outside the tube with distance b be P.
Consider an Amperian loop, as shown in the figure.
Length of the loop, l = 2π×b=2πb
Current enclosed in the loop = i
On applying Ampere's law, we get
B.dl=μ0iB×2πb=μ0iB=μ0i2πb

Page No 252:

Question 49:




a) The magnetic field inside any conducting tube is always zero.
∴ Magnetic field just inside the tube is zero.

(b) Let the point outside the tube with distance b be P.
Consider an Amperian loop, as shown in the figure.
Length of the loop, l = 2π×b=2πb
Current enclosed in the loop = i
On applying Ampere's law, we get
B.dl=μ0iB×2πb=μ0iB=μ0i2πb

Answer:

Given:
Magnitude of current = i
Radius of the wire = b


For a point at a distance a from the axis,
Current enclosed, i'=iπb2×πa2
By Ampere's circuital law,
B.dl = μ0i'
For the given conditions,
B×2πa= μ0iπb2×πa2B=μ0ia2πb2

Page No 252:

Question 50:

Given:
Magnitude of current = i
Radius of the wire = b


For a point at a distance a from the axis,
Current enclosed, i'=iπb2×πa2
By Ampere's circuital law,
B.dl = μ0i'
For the given conditions,
B×2πa= μ0iπb2×πa2B=μ0ia2πb2

Answer:


Given:
Magnitude of current, i = 5 A
Radius of the wire, b=10 cm=10×10-2 m
For a point at a distance a from the axis,
Current enclosed, i' = iπb2×πa2
By Ampere's circuital law,
B.dl = μ0i'
For the given conditions,
B×2πa = μ0iπb2×πa2B = μ0ia2πb2    1
(a) a=2 cm=2×10-2 m
Again, using the circuital law, we get
B = 4π×10-7×5×2×10-22π×10-2= 2×10-6 T=2 μT

(b) On putting a = 10 cm = 10×10-2 m in (1), we get
B = 10 μT

(c)Using the circuital law, we get
 B.dl = μ0iB = μ0i2πa = 2×10-7×520×10-2= 5×10-6 T = 5 μT

Page No 252:

Question 51:


Given:
Magnitude of current, i = 5 A
Radius of the wire, b=10 cm=10×10-2 m
For a point at a distance a from the axis,
Current enclosed, i' = iπb2×πa2
By Ampere's circuital law,
B.dl = μ0i'
For the given conditions,
B×2πa = μ0iπb2×πa2B = μ0ia2πb2    1
(a) a=2 cm=2×10-2 m
Again, using the circuital law, we get
B = 4π×10-7×5×2×10-22π×10-2= 2×10-6 T=2 μT

(b) On putting a = 10 cm = 10×10-2 m in (1), we get
B = 10 μT

(c)Using the circuital law, we get
 B.dl = μ0iB = μ0i2πa = 2×10-7×520×10-2= 5×10-6 T = 5 μT

Answer:

Half of the loop PQRS is in the region of magnetic field and half in the area of zero magnetic field.

Let us consider a current carrying circular wire, due to which there is uniform magnetic field in the region.

Take a point A inside the loop PQRS in the region where B = 0

According to Ampere's circuital law,
B.dl=μ0i

If there is current enclosed by the loop PQRS, then magnetic field B cannot be 0.

Whereas, we have taken the magnetic field at point A to be zero.

Thus, such a field is not possible.

Page No 252:

Question 52:

Half of the loop PQRS is in the region of magnetic field and half in the area of zero magnetic field.

Let us consider a current carrying circular wire, due to which there is uniform magnetic field in the region.

Take a point A inside the loop PQRS in the region where B = 0

According to Ampere's circuital law,
B.dl=μ0i

If there is current enclosed by the loop PQRS, then magnetic field B cannot be 0.

Whereas, we have taken the magnetic field at point A to be zero.

Thus, such a field is not possible.

Answer:

At point P,
Current, i = 0
∴ Magnetic field B = 0
At point Q,
Applying Ampere's law, we get
B.dl=μ0i B.dl=μ0kdl B=μ0k
At point R,
Current, i = 0
∴ B = 0

Page No 252:

Question 53:

At point P,
Current, i = 0
∴ Magnetic field B = 0
At point Q,
Applying Ampere's law, we get
B.dl=μ0i B.dl=μ0kdl B=μ0k
At point R,
Current, i = 0
∴ B = 0

Answer:

Given: 
Charge = q
Mass = m
Radius = r
We know that the radius described by a charged particle in a magnetic field is given by
r=mvqB
Using Ampere circuital lawB.dl=μ0i B.dl=μ0kdl B=μ0k
v=Bqrm=μ0kqrm

Page No 252:

Question 54:

Given: 
Charge = q
Mass = m
Radius = r
We know that the radius described by a charged particle in a magnetic field is given by
r=mvqB
Using Ampere circuital lawB.dl=μ0i B.dl=μ0kdl B=μ0k
v=Bqrm=μ0kqrm

Answer:

Given:
Magnitude of current, i = 5 A
Magnetic field intensity, B = 3.14 × 10−2 T
We know that the magnetic field inside a long solenoid having n turns per unit length is given by
B=μ0ni3.14×10-2 =4π×10-7×n×5n=10-220×10-7=5×103=5000 turns/m

Page No 252:

Question 55:

Given:
Magnitude of current, i = 5 A
Magnetic field intensity, B = 3.14 × 10−2 T
We know that the magnetic field inside a long solenoid having n turns per unit length is given by
B=μ0ni3.14×10-2 =4π×10-7×n×5n=10-220×10-7=5×103=5000 turns/m

Answer:

Given:
Radius of the wire, r = 0.5 mm
Width of each turn, = diameter of the wire, 2r = 1 mm = 1 × 10−3 m
∴ Total number of turns in 1 m solenoid
n=11×10-3=103Magnitude of current, i=5 AUsing B=μ0ni, we getB=4π×10-7×103×5B=2π×10-3 T

Page No 252:

Question 56:

Given:
Radius of the wire, r = 0.5 mm
Width of each turn, = diameter of the wire, 2r = 1 mm = 1 × 10−3 m
∴ Total number of turns in 1 m solenoid
n=11×10-3=103Magnitude of current, i=5 AUsing B=μ0ni, we getB=4π×10-7×103×5B=2π×10-3 T

Answer:

Given:
Resistance per unit length of the wire, Rl = 0.01 Ω/m
Radius of the wire, r = 1.0 cm = 0.01 m
Total no. of turns, N = 400
Magnetic field intensity, B = 1.0 × 10−2T
Now,
Let E be the emf of the battery and R0 be the total resistance of the wire.
i=ER0=E0.01×2πr×400=E0.01×2×π×0.01×400

The magnetic field near the centre of the solenoid is given by
B=μ0ni=1×10-2=4π×10-7×40020×10-2×E2π×4×10-2E=10-2×20×10-2×2×10-210-7×4×102=1 V



Page No 253:

Question 57:

Given:
Resistance per unit length of the wire, Rl = 0.01 Ω/m
Radius of the wire, r = 1.0 cm = 0.01 m
Total no. of turns, N = 400
Magnetic field intensity, B = 1.0 × 10−2T
Now,
Let E be the emf of the battery and R0 be the total resistance of the wire.
i=ER0=E0.01×2πr×400=E0.01×2×π×0.01×400

The magnetic field near the centre of the solenoid is given by
B=μ0ni=1×10-2=4π×10-7×40020×10-2×E2π×4×10-2E=10-2×20×10-2×2×10-210-7×4×102=1 V

Answer:

(a) Given:
Current in the loop or circular current = indx
Radius of the loop having circular current  = r
Distance of the centre of the solenoid from the circular current = l2-x
Magnetic field at the centre due to the circular loop,
B=μ02ir2(x2+r2)3/2
B=dB=01μ0a2nidx4πa2+(l-2x)23/2=01μ0nia2dx4πa31+l-2xa23/2=μ0ni4πa01dx1+l-2xa23/2=μ0ni4πa.4πa1+2al=μ0ni1+2al2
(b) When a > > l,
B=μ0ni2a

Page No 253:

Question 58:

(a) Given:
Current in the loop or circular current = indx
Radius of the loop having circular current  = r
Distance of the centre of the solenoid from the circular current = l2-x
Magnetic field at the centre due to the circular loop,
B=μ02ir2(x2+r2)3/2
B=dB=01μ0a2nidx4πa2+(l-2x)23/2=01μ0nia2dx4πa31+l-2xa23/2=μ0ni4πa01dx1+l-2xa23/2=μ0ni4πa.4πa1+2al=μ0ni1+2al2
(b) When a > > l,
B=μ0ni2a

Answer:

Given:
Magnitude of current in the solenoid, i = 2 A
Frequency of the electron, f=1×108 rev/s
Mass of the electron, m=9.1×10-31 kg
Charge of the electron, q=1.6×10-19 C
We know that the magnetic field inside a solenoid is given by
B = µ0ni
If a particle executes uniform circular motion inside a magnetic field, the frequency of the particle is given by
f=qB2πmB=2πmfq μ0ni=2πmfq   [Using (1)]n=2πmfμ0qi=2π×9.1×10-31×1×1084π×10-7×1.6×10-19×2=1420 turns/m

Page No 253:

Question 59:

Given:
Magnitude of current in the solenoid, i = 2 A
Frequency of the electron, f=1×108 rev/s
Mass of the electron, m=9.1×10-31 kg
Charge of the electron, q=1.6×10-19 C
We know that the magnetic field inside a solenoid is given by
B = µ0ni
If a particle executes uniform circular motion inside a magnetic field, the frequency of the particle is given by
f=qB2πmB=2πmfq μ0ni=2πmfq   [Using (1)]n=2πmfμ0qi=2π×9.1×10-31×1×1084π×10-7×1.6×10-19×2=1420 turns/m

Answer:

Given:
Magnitude of current in the solenoid = i
Number of turns per unit length = n
When a particle is projected perpendicular to the magnetic field, it describes a circular path.
And for the particle (projected from a point on the axis in a direction perpendicular to the axis) to not strike the solenoid, the maximum radius of that circular path should be r/2.

∴ Radius of the circle = r2

We know,
Centripetal force = Magnetic force
mV2r=qVBV=qBrm=qμ0nir2m

Page No 253:

Question 60:

Given:
Magnitude of current in the solenoid = i
Number of turns per unit length = n
When a particle is projected perpendicular to the magnetic field, it describes a circular path.
And for the particle (projected from a point on the axis in a direction perpendicular to the axis) to not strike the solenoid, the maximum radius of that circular path should be r/2.

∴ Radius of the circle = r2

We know,
Centripetal force = Magnetic force
mV2r=qVBV=qBrm=qμ0nir2m

Answer:

Given:
Number of turns per unit length of the solenoid = n
(a) Since the net magnetic field near the centre of the solenoid is 0,
B plate=B solenoidUsing Guass's law for the plate, we getBplate×2l=μ0i=μ0klBplate=μ0k2    1

For the solenoid,
Bsolenoid=μ0ni    ...(2)

From (1) and (2), we get
i = k2n

(b) On putting the value of n in (2), we get
B solenoid=μ0k2
Now,  B plate and B solenoid are perpendicular to each other.
Thus, the net magnetic field near the centre of the solenoid is given by
Bnet = μ0k22+μ0k22       =μ0k2

Page No 253:

Question 61:

Given:
Number of turns per unit length of the solenoid = n
(a) Since the net magnetic field near the centre of the solenoid is 0,
B plate=B solenoidUsing Guass's law for the plate, we getBplate×2l=μ0i=μ0klBplate=μ0k2    1

For the solenoid,
Bsolenoid=μ0ni    ...(2)

From (1) and (2), we get
i = k2n

(b) On putting the value of n in (2), we get
B solenoid=μ0k2
Now,  B plate and B solenoid are perpendicular to each other.
Thus, the net magnetic field near the centre of the solenoid is given by
Bnet = μ0k22+μ0k22       =μ0k2

Answer:

Given:
Capacitance, C = 100 microfarad
Voltage, V = 20 V
Charge stored in the capacitor, Q = CV
= 100×10-6×20= 2×10-3 C
It is given that the potential difference across the capacitor drops to 90% of its maximum value.
Thus,
V'=90100×20=18 VNew charge, Q'=CV'=1.8×10-3 CNow, Current, i=Q-Q'ti=(2-1.8)×10-32=2×10-42i=1×10-4 A

No. of turns per metre, n = 4000
Thus, the average magnetic field at the centre of the solenoid is given by
B=μ0ni=4π×10-7×4000×10-4=16π×10-8 T



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