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Page No 22:

Question 1:

Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field B = B0k^.

(a) They have equal z-components of momenta.
(b) They must have equal charges.
(c) They necessarily represent a particle-antiparticle pair.
(d) The charge to mass ratio satisfy : em1+em2=0.


Since path of the charge particle is helix, it means it is thrown in the field B at some angle θ and its component of velocity along the field be v cosθ and perpendicular to the field be v sinθ.
So, the pitch of the helix will be,
p = Time period × v cosθ
  =2πmqB×v cosθ    T=2πmqB
  qm=2πpB×v cosθ
For a given pitch, qm must be constant since, the two charged particles traverse identical helical paths (same pitch) in completely opposite sense (i.e. nature of charge must be opposite) in the uniform magnetic field B, so, qm for the two particles must be same but of opposite sign.
So, q1m1+q2m2=0
q1=e1 and q2=e2  Then the given expression can be written as:
Hence, the correct answer is option (d).

Page No 22:

Question 2:

Biot-Savart law indicates that the moving electrons (velocity vproduce a magnetic field B such that
(a) B v.
(b) B || v.
(c) it obeys inverse cube law.
(d) it is along the line joining the electron and point of observation.


According to Biot-Savart law:


Here, q is charge, v is velocity, r is position vector.

B=μ4π×q v r sinθr3=μ4π×q v sinθr2

(Here, θ is angle between v and r)
So, in the given options, the moving charge will produce magnetic field if:
θ=90°, i.e., Bv.
But if B||v, then θ=0 B=0
Hence, the correct answer is option (a).

Page No 23:

Question 3:

A current carrying circular loop of radius R is placed in the x-y plane with centre at the origin. Half of the loop with x > 0 is now bent so that it now lies in the y-z plane.
(a) The magnitude of magnetic moment now diminishes.
(b) The magnetic moment does not change.
(c) The magnitude of B at (0.0.z), z >>R increases.
(d) The magnitude of B at (0.0.z), z >>R is unchanged.


When the loop of radius R is in x-y plane then magnetic moment,

m1=I×A1 = IπR2 (along z-direction)

Now, when the loop is bent that one half in y-z plane, then magnetic moment,

m2=I×πR22 along z-axis + I×πR22 along x-axis

So, the effective magnetic moment due to entire loop when bent,



Hence, the correct answer is option (a).

Page No 23:

Question 4:

An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true?
(a) The electron will be accelerated along the axis.
(b) The electron path will be circular about the axis.
(c) The electron will experience a force at 45° to the axis and hence execute a helical path.
(d) The electron will continue to move with uniform velocity along the axis of the solenoid.


If an electron (e) is projected with velocity v along the axis of a solenoid (having uniform magnetic field B along its axis), the magnetic force of the electron,

F=ev×B=e v B sinθ

But, θ = 0° (as v is along B)

So, F = e v B × 0 = 0

Hence, the electron will continue to move with uniform velocity along the axis of the solenoid.
Hence, the correct answer is option (d).

Page No 23:

Question 5:

In a cyclotron, a charged particle
(a) undergoes acceleration all the time.
(b) speeds up between the dees because of the magnetic field.
(c) speeds up in a dee.
(d) slows down within a dee and speeds up between dees.


Cyclotron is a device, which is used to accelerate the charged particles. In cyclotron, charged particles accelerates all the time.
Hence, the correct answer is option (a).

Page No 23:

Question 6:

A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field B. The work done to rotate the loop by 30° about an axis perpendicular to its plane is
(a) MB.

(b) 3MB2.

(c) MB2.

(d) zero.


According to the question, the loop should be rotated about an axis perpendicular to its plane. It means, the angle between area vector and the field will always zero. So, the work done to rotate the loop is zero.
Hence, the correct answer is option (d).

Page No 23:

Question 7:

The gyro-magnetic ratio of an electron in an H-atom, according to Bohr model, is
(a) independent of which orbit it is in.
(b) negative.
(c) positive.
(d) increases with the quantum number n.


For, an electron in nth orbit of hydrogen atom,

Angular momentum, Ln h2π

Magnetic moment, Mn e h4πm
Gyromagnetic ratio = ML=n e h4πmn h2π

This expression is independent of n, i.e., gyromagnetic ratio is independent of the orbit in which the electron is revolving.
Now, charge on an electron is negative so, this ratio is negative.
Hence, the correct answers are options (a) and (b).

Page No 24:

Question 8:

Consider a wire carrying a steady current, I placed in a uniform magnetic field B perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that,
(a) motion of charges inside the conductor is unaffected by B since they do not absorb energy.
(b) some charges inside the wire move to the surface as a result of B.
(c) if the wire moves under the influence of B, no work is done by the force.
(d) if the wire moves under the influence of B, no work is done by the magnetic force on the ions, assumed fixed within the wire.


According to question, the wire is perpendicular to the field. Magnetic force on a moving charge in a uniform magnetic field:

F=q v×BF=q v B sinθ

This force may cause the charge particles to move to the surface of the wire. Now, if the wire is moving, but the charge particles inside the wire are fixed, then the magnetic force will be perpendicular to the displacement of the ion, and hence, there will be no work done.
Hence, the correct answers are options (b) and (d).


Page No 24:

Question 9:

Two identical current carrying coaxial loops, carry current I in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C,
(a)  .cB.dl=2μ0I.
(b) the value of cB.dl is independent of sense of C.
(c) there may be a point on C where B and dl are perpendicular.
(d) B vanishes everywhere on C.


Both the current loops having current flowing in opposite directions. Now if we draw an amperian loop (C) including both the loops once then, total current within the loop


(∵ currents are in opposite direction.)
Now, according to Ampere's law

 B·d l=μi=0

Which is independent of sense of C.
It also indicates that, there may be a point on C where, B and dl are perpendicular for which angle between them be 90° and hence, B·dl=0
Hence, the correct answers are options (b) and (c).

Page No 24:

Question 10:

A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity v and a positron enters via opposite face with velocity - v. At this instant,
(a) the electric forces on both the particles cause identical accelerations.
(b) the magnetic forces on both the particles cause equal accelerations.
(c) both particles gain or loose energy at the same rate.
(d) the motion of the centre of mass (CM) is determined by B alone.


E=electric fieldFE=force due to electric fieldFm=magnetic forceB=magnetic field

Since, magnitude of charge and mass on an electron (e) and positron (e+) is same, then the magnitude of electrostatic force (FE) and magnetic force (Fm) are,
FE = e E (are same for both electron and positron)
Fm = e v B (are also same for both electron and same)
But the nature of charge on electron is negative (–e) and on positron is positive (+e), so the acceleration caused due to electrostatic force are opposite to each-other as shown in the diagram.
As, shown in the diagram magnetic force on both the charged particles are equal and in the same direction. So, the acceleration caused due to magnetic field on both the particles are equal, i.e., evBm. Also, both the particles will lose energy at same rate.
Now, the net force on the positron and electron pair will be:

2 Fm in downward direction
So, the motion of centre of mass (COM) will be determined by magnetic field B.
Hence, the correct answers are options b, c, and d.

Page No 24:

Question 11:

A charged particle would continue to move with a constant velocity in a region wherein,
(a) E = 0, B  0.
(b) ≠ 0, B  0.
(c) E  0, B = 0.
(d) E = 0, B = 0.


Let the charge on the particle be q and it's velocity be v, which is constant. Now, if E=0 and B0 then, electrostatic force will not act and only magnetic force will act. Due to which the particle will move in a circular path and centripetal force will be governed by the magnetic force, Fm = qvB which acts towards the centre of the circular path. The magnitude of the velocity will remain constant.
If E0 and B0, and the direction of electric and magnetic field is such that the net force on the charged particle is balanced, then also the magnitude of velocity remains constant.
If E0, B=0, then electrostatic force will cause the change in velocity of the charged particle.
If E=0 and B=0, then neither electrostatic nor magnetic force will act on the charged particle. So, velocity will remain constant.
Hence, the correct answers are options a, b and d.

Page No 25:

Question 12:

Verify that the cyclotron frequency ω = eB/m has the correct dimensions of [T]–1.


If a charged particle (e) is moving perpendicular to the magnetic field (B) with velocity (v), then centripetal force,

Fc=mv2R=qvBqBm=vRqBm=ω (v=ω×R)

So, dimensionally

Page No 25:

Question 13:

Show that a force that does no work must be a velocity dependent force.


Work done,

W=F·dxIf, W =F·dx=0F·vdt=0     ( velocity v=dxdt)F·v=0

⇒ Angle between force and velocity is 90°.
Hence, force F must be velocity dependent. If v changes (direction) then (directions) F should also change such that above condition is satisfied.

Page No 25:

Question 14:

The magnetic force depends on v which depends on the inertial frame of reference. Does then the magnetic force differ from inertial frame to frame? Is it reasonable that the net acceleration has a different value in different frames of reference?


Magnetic force on a charged particle (q) moving with velocity, v in an inertial frame in magnetic field region B,

Fm=qv×B=q v B sinθ

(θ = angle between the velocity and B)
So, the magnetic force on a charged particle will differ from one inertial frame to another frame of reference i.e., magnetic force is frame dependent. But, net acceleration arising from this force is frame independent for inertial frames.

Page No 25:

Question 15:

Describe the motion of a charged particle in a cyclotron if the frequency of the radio frequency (rf) field were doubled.


Time period of the charged particle (q)

T = 2πmqB  (q = charge, B = magnetic field, m = mass of the particle)

Frequency, f1T=qB2πm

Now, if frequency is increased double,

i.e., f'=2f

Then, its time period, T'=1f'=12f=T2=πmqB

So, it can be said that the particle will accelerate but the radius of the path will remain unchanged.

Page No 25:

Question 16:

Two long wires carrying current I1 and I2 are arranged as shown in the given figure. The one carrying current I1 is along is the x-axis. The other carrying current I2 is along a line parallel to the y-axis given by x = 0 and z = d. Find the force exerted at O2 because of the wire along the x-axis.


Magnetic field due to current carrying wire (I1) at O2 be,

B12=μI12πd   (along y-axis)

But the second wire carrying current (I2) is along y-axis. So, magnetic force on the second wire carrying current (I2) at Owill be,
F=I2dl×B12  =0    (∵ angle between dl and B12 is zero, i.e. both are in y-direction)

Page No 25:

Question 17:

A current carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-z and z-x planes with their centres at the origin, joined together. Find the direction and magnitude of B at the origin.


As shown in the diagram three quarter circles in y, z and x planes.

The magnetic field due the quarter current carrying loop in – y plane is:
B1=μi2R×90°360°=μi8R  k^ 

Similarly, for the loop in yplane:

B2=μi8R i^

For the loop in z-x plane:

B3=μi8R j^

So, the resultant magnetic field on the completed loop will be,

B=B1+B2+B3=μi8R i^ + μi8R j^ + μi8R k^  =14 i^+j^+k^  μi2R

Page No 25:

Question 18:

A charged particle of charge e and mass m is moving in an electric field E and magnetic field B. Construct dimensionless quantities and quantities of dimension [T ]–1.


When a charged particle enters in a region where electric field and magnetic field are perpendicular, then the charged particle will follow the helical path and then, its time period,

T = 2πrv=2πω    ω=vr

Here, r is the radius and  ω is angular velocity.


Page No 25:

Question 19:

An electron enters with a velocity v = v0i into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in plane parallel to the x-y plane. Suggest a configuration of fields E and B that can lead to it.


Since, the orbit of the electron spiral down inside the cube in the plane parallel to xy plane, so

Magnetic field B must be along +z-axis and electric field E must be along +x-axis.

Because, Fm=q v×B will try to move the particle in circular path but, electrostatic force,

Fe=q E will act in x-direction.



Such that, E, B>0

This condition will ensure the movement of the charged particle in spiral path parallel to the x-y plane.

Page No 25:

Question 20:

Do magnetic forces obey Newton’s  third law. Verify  for two current elements dl1 = dl i^ located at the origin and dl2 = dl j^ located at (0, R, 0).  Both carry current I.


Given, dl1=dli^
and, dl2=dlj^

Now, magnetic field at current element dl2 due to current element dl1,

B12=μI dl4πR2 k^

So, force on current element dl2 be,

F12=I dl j^×B12=I dl j^×μI dl4πR2 k^                        =μI2dl24πR2 l^        j^×k^=i^

Which is non zero,
But, there is no effect of magnetic field of current element dl2 on current element dl1

So, F21=0

it means, F12F21

Which violates the Newton's third law.

Page No 26:

Question 21:

A multirange voltmeter can be constructed by using a galvanometer circuit as shown in the given figure. We want to construct a voltmeter that can measure 2V, 20V and 200V using a galvanometer of resistance 10Ω and that produces maximum deflection for current of 1 mA. Find R1, R2 and R3 that have to be used.


Resistance of the galvanometer, R = 10 Ω
It will produce maximum deflection, when current,
I = 1 mA = 1 × 10–3 A
for, 2 V range,
I (R + R1) = 2

R+R1=2IR1=2I-RR1=21×10-3-10=1990 Ω

For, 20 V range,
I (R + R1 + R2) = 20

R2=20I-R-R1        =201×10-3-10-1990=18000 Ω

For, 200 V range,
I (R + R1 + RR3) = 200

​R3=200I-R-R1-R2        = 2001×10-3-10-1990-18000        =180000 Ω

Page No 26:

Question 22:

A long straight wire carrying current of 25A rests on a table as shown in the given figure. Another wire PQ of length 1m, mass 2.5 g carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?



Current through the straight long wire,
I1 = 25 A        (right to left)
Current through PQ,
I2 = 25 A        (left to right)
i.e., I1 = I2 = I
Length of PQ,
l = 1 m
Mass of the wire
m = 2.5 g = 2.5 × 10–3 kg
Let, the wire PQ rises to the height h.
Gravitational force on the wire be,
FG = mg
Magnetic force on the wire PQ due the long straight wire,


At, height h,
FG = Fm


h=μI2l2πmg=4π×10-7×252×12π×2.5×10-3×9.8                 μ=4π×10-7, g=9.8 m/s2h=51.02×10-4 m      =0.51 cm
Hence, the required height will be 0.51 cm.

Page No 26:

Question 23:

A 100 turn rectangular coil ABCD (in XY plane) is hung from one arm of a balance in the given figure. A mass 500g is added to the other arm to balance the weight of the coil. A current 4.9 A passes through the coil and a constant magnetic field of 0.2 T acting inward (in xz plane) is switched on such that only arm CD of length 1 cm lies in the field. How much additional mass ‘m’ must be added to regain the balance?



Mass, m= 500 g = 500 × 10–3 kg,
Magnetic field, B = 0.2 T,
Current, I = 4.9 A and 
CD = 1 cm 
Initially, when there is no magnetic field, then net torque about the pivot must be zero, i.e., 
So, m1g × = Wcoil ×
⇒ Wcoil = m1g
Weight of the coil = m1g          .....(1)
Now, when the magnetic field is switched on and additional mass m is added, then magnetic force (IlB) starts acting on the coil in downward direction:
So, (m+ m)glWcoil × l + (IlB) ×
⇒ mgl = Wcoil × l – m1gl + (IlB)l
            = 0 + (IlB)         (∵ Wcoil = m1g)
m=IlBlgl        =IBlg        =4.9×0.2×1×10-29.8m=1×10-3 kg
Hence, the required additional mass will be 10–3 kg.

Page No 26:

Question 24:

A rectangular conducting loop consists of two wires on two opposite sides of length l joined together by rods of length d. The wires are each of the same material but with cross-sections differing by a factor of 2. The thicker wire has a resistance R and the rods are of low resistance, which in turn are connected to a constant voltage source V0. The loop is placed in uniform a magnetic field B at 45° to its plane. Find τ, the torque exerted by the magnetic field on the loop about an axis through the centres of rods.


Given length of the parallel wires is l and distance between them is d.
Let the current through the thicker and thinner wire be I1 and I2 respectively.
Let the resistance of the thicker wire be R1 = R
But, R2=ρlA2
So, resistance of the thinner wire will be:
R2=ρlA2=ρlA12=2R            area, A1A2=2
Now, magnetic force on the thicker wire be, 
F1 = I1lB           (∵ loop is connected to the voltage source of voltage Vο)
     =VοR1lB         Vο=I1R1=VοRlB
But the magnetic field is in the direction of 45º to the plane of the loop.
So, torque about the centre of the rods
τ1=d22 F1=d22×VοRlB
Net torque,
τ1-τ2=d22×VοRlB-d22×Vο2RlB         =142×VοlBR


Page No 26:

Question 25:

An electron and a positron are released from (0, 0, 0) and (0, 0, 1.5R ) respectively, in a uniform magnetic field B = B0i^, each with an equal momentum of magnitude p = e BR. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?


Here, p1 and p2 are the momentum of electron(e) and positron (e+).
pand p2 makes an angle θ with y-axis.

Let, both electron and positron traverse a circle of radius R.
Let Cand Cbe the centre line of their respective circular path.
So, the co-ordinates of Cand Cp can be given as,

C(0, – R sin θ, R cos θ)
Cp(0, + R sin θ, 1.5 R – Rcos θ)

Now, if the distance between the centres of the two circular paths is more than 2R i.e., more than the diameter of either circle, then the two circles will never overlap.
Distance between the centres,
D=0-02+-R sin θ-R sin θ2+R cos θ-1.5R+R cos θ2
D2 = 4R2 + 2.25 R2 – 6Rcos θ 
But for not overlapping the circles 
> 2R
i.e. D> 4R2
So, 4R2+94R26R2 cos θ>4R2
94R2>6R2 cos θcos θ<38
This is the required condition.

Page No 26:

Question 26:

A uniform conducting wire of length 12a and resistance R is wound up as a current carrying coil in the shape of (i) an equilateral triangle of side a; (ii) a square of sides a and, (iii) a regular hexagon of sides a. The coil is connected to a voltage source V0. Find the magnetic moment of the coils in each case.


Total length of the conducting wire, l = 12a
Let the current flowing in each loop be I.
(i) So, number of loops, if the wire is wound in the form of an equilateral triangle of side a,

So, area of the triangle
Magnetic moment, 
m1 = n1(IA1=4×I×34a2=3Ia2           .....1

(ii) Now for square of side a
Number of loops,

Area of the square loops,
A2 = a2
Magnetic moments,
m2 = n2(IA2)
     = 3 × I × a2
     = 3Ia2          
a2=a2m2=3Ia2         ....(2)
(iii) Now, for regular hexagon loop:
Number of loops,
Area of hexagonal loops, 
Magnetic moments,
m3 = n3(IA3)

Page No 27:

Question 27:

Consider a circular current-carrying loop of radius R in the x-y plane with centre at origin. Consider the line intergral
IL=-LLB.dl taken along zaxis.
(a) Show that (L) monotonically increases with L.
(b) Use an appropriate Amperian loop to show that ℑ (∞) = μ0Iwhere I is the current in the wire.
(c) Verify directly the above result.
(d) Suppose we replace the circular coil by a square coil of sides R carrying the same current I. What can you say about (L) and() ?


Consider the circular loop of radius R in x-y plane, such that its axis passes through its centre 0 along z-axis.

Consider two points at a distance +L and –L along the z-axis from the centre of the loop 0.
(a) As shown in the diagram, the sense of magnetic field due to the current carrying loop is along z-axis.
The angle between B and dl is 0º.
So,  ​IL=-LLB.dl=-LLBdl cos 0º
It means, its value varies monotonically with L.
(b) Now assume an amperian loop MPNM as shown in the diagram.
So, for IL at L
I +0=μο I
Because contribution from large distance on contour → 0  as, B1r3
So, I=μο I

(c) Now, magnetic field due to the circular loops along the z-axis, at a distance r from the origin 
For, large distance 
Let, r = R tan θ, then dr = R(secθ)θ
So, I=μοI2-π2π2cos θ d θ              =μοI2sin θ-π2π2              =μοI2sinπ2-sin-π2              =μοI2×1+1I=μοI

(d) Now, if we replace the circular coil with the square coil then magnetic field,
Bz​(square) < Bz(circular coil)
I(L)square<I(L)circular coil
But for large distance, 
I()square<I()circular coil
[The same is proved in part (b)]

Page No 27:

Question 28:

A multirange current meter can be constructed by using a galvanometer circuit as shown in the given figure. We want a current meter that can measure 10mA, 100mA and 1A using a galvanometer of resistance 10Ω and that prduces maximum deflection for current of 1mA. Find S1, S2 and S3 that have to be used.


Resistance of the galvanometer, 
R = 10 Ω
The maximum deflection for current 
I = 1 mA = 1 × 10–3 A

Now, for I= 10 mA current,
I × R = (I1I)(S1S2 + S3)          .....(1)
For, I= 100 mA current,
I × (R+S1) = (I2 – I)(S2 + S3)                 .....(2)
and for, I3 = 1 A current
I ×
(R+S1+S2) = (I3 I) (S3)                        .....(3)
Now, by putting the values of I, I1, I2, I3 and and solving the equations (1), (2) and (3) we get:
S1 = 1 
S2 = 0.1 Ω
S3 = 0.01 Ω​


Page No 27:

Question 29:

Five long wires A, B, C, D and E, each carrying current I are arranged to form edges of a pentagonal prism as shown in the given figure. Each carries current out of the plane of paper.
(a) What will be magnetic induction at a point on the axis O? Axis is at a distance R from each wire.
(b) What will be the field if current in one of the wires (say A) is switched off?
(c) What if current in one of the wire (say) A is reversed?


(a) Since, each wire carries equal current (I), and point O is at equal distance R from each wire.
Hence, due to symmetry of the shape, the net field at point O will be zero.
BA+BB+Bc+BD +BE = 0        .....(1)
i.e. Bnet= 0  
(b) Now if the current in one of the wire (let for A) is switched off,
then, resultant of remaining fields will be.
BA+BB+Bc+BD +BE = 0        from (1)  BB+Bc+BD+BE=-BABB+Bc+BD+BE=-μοI2πR=μοI2πR 
But the direction must be opposite.
Now, if the current in a wire (let for A) is reversed then magnetic field due to that wire will be in opposite direction but of same magnitude as shown in figure. Now using vector sum of magnetic fields due to currents at point A, B, C, D and E we, get 
​Bnet=μοIπR perpendicular to AO towards left.

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