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#### Page No 275:

#### Question 1:

#### Answer:

No, we cannot have a single north or south pole. Magnetic poles are always found in pairs. They are equal in strength and opposite in nature. Even if we break a magnet into a number of pieces, each piece will become a magnet with equal and opposite poles.

#### Page No 275:

#### Question 2:

No, we cannot have a single north or south pole. Magnetic poles are always found in pairs. They are equal in strength and opposite in nature. Even if we break a magnet into a number of pieces, each piece will become a magnet with equal and opposite poles.

#### Answer:

No, two distinct poles cannot exist at two nearby points in a magnet, as a magnet contains only two distinct poles located at its ends.

#### Page No 275:

#### Question 3:

No, two distinct poles cannot exist at two nearby points in a magnet, as a magnet contains only two distinct poles located at its ends.

#### Answer:

No, the material making up the middle region of a magnet is the same as that of the material making up its end. When an iron needle is taken closer to one of the ends of a magnet, the pole of the magnet induces a pole of opposite polarity on the needle, making the needle a magnet itself and thereby making it attracted to that pole.

But if we bring the needle closer to the centre of the magnet, then both the poles of the magnet will induce opposite polarity on the needle. As a result, the needle will not get attracted towards the centre of the magnet.

#### Page No 275:

#### Question 4:

No, the material making up the middle region of a magnet is the same as that of the material making up its end. When an iron needle is taken closer to one of the ends of a magnet, the pole of the magnet induces a pole of opposite polarity on the needle, making the needle a magnet itself and thereby making it attracted to that pole.

But if we bring the needle closer to the centre of the magnet, then both the poles of the magnet will induce opposite polarity on the needle. As a result, the needle will not get attracted towards the centre of the magnet.

#### Answer:

The direction of the magnetic field is the same in both cases, that is, inside a solenoid and inside a bar magnet. In a solenoid, magnetic field lines are directed from one end to the other internally and externally, so they are in the equivalent combination of north and south poles (as shown in figure).

#### Page No 275:

#### Question 5:

The direction of the magnetic field is the same in both cases, that is, inside a solenoid and inside a bar magnet. In a solenoid, magnetic field lines are directed from one end to the other internally and externally, so they are in the equivalent combination of north and south poles (as shown in figure).

#### Answer:

The difference between the two configurations is that in the current-carrying loop, the magnetic field lines pass through the centre and are perpendicular to its axis; whereas in the equivalent magnetic dipole, the magnetic field lines do not pass through the centre.

#### Page No 275:

#### Question 6:

The difference between the two configurations is that in the current-carrying loop, the magnetic field lines pass through the centre and are perpendicular to its axis; whereas in the equivalent magnetic dipole, the magnetic field lines do not pass through the centre.

#### Answer:

Yes, it seems to contradict with our earlier knowledge that a magnetic field can exert forces only perpendicular to itself.

$\overrightarrow{F}=m\overrightarrow{B}$

Here,

$\overrightarrow{B}$ = Magnetic field*m* = Magnetic charge

For a positive magnetic charge, force is along the magnetic field.

For a negative magnetic charge, force is opposite to the magnetic field.

Thus, it contradicts the notion that a magnetic field can exert forces only perpendicular to itself.

#### Page No 275:

#### Question 7:

Yes, it seems to contradict with our earlier knowledge that a magnetic field can exert forces only perpendicular to itself.

$\overrightarrow{F}=m\overrightarrow{B}$

Here,

$\overrightarrow{B}$ = Magnetic field*m* = Magnetic charge

For a positive magnetic charge, force is along the magnetic field.

For a negative magnetic charge, force is opposite to the magnetic field.

Thus, it contradicts the notion that a magnetic field can exert forces only perpendicular to itself.

#### Answer:

Yes, it contradicts our earlier knowledge that magnetic forces cannot do any work and hence cannot increase the kinetic energy of the system. When opposite poles are facing each other, an attractive force acts between them so the magnets are pulled towards each other. As the two magnets come close to each other so the force between them increases and hence, the kinetic energy also increases.

#### Page No 275:

#### Question 8:

Yes, it contradicts our earlier knowledge that magnetic forces cannot do any work and hence cannot increase the kinetic energy of the system. When opposite poles are facing each other, an attractive force acts between them so the magnets are pulled towards each other. As the two magnets come close to each other so the force between them increases and hence, the kinetic energy also increases.

#### Answer:

No, we cannot have a magnetic scalar potential here.

Ampere's law is a method of calculating magnetic field due to current distribution. On the other hand, magnetic scalar potential requires a magnetic field due to pole strength *m*.

Potential at a distance *r* is given by

$\frac{{\mu}_{0}m}{4\mathrm{\pi}r}$

As there is no current distribution, no magnetic field due to poles or the pole strength is present. That is why we cannot have a magnetic scalar potential in this case.

#### Page No 275:

#### Question 9:

No, we cannot have a magnetic scalar potential here.

Ampere's law is a method of calculating magnetic field due to current distribution. On the other hand, magnetic scalar potential requires a magnetic field due to pole strength *m*.

Potential at a distance *r* is given by

$\frac{{\mu}_{0}m}{4\mathrm{\pi}r}$

As there is no current distribution, no magnetic field due to poles or the pole strength is present. That is why we cannot have a magnetic scalar potential in this case.

#### Answer:

Yes, Earth's magnetic field is vertical at the poles. A freely suspended magnet becomes vertical at the poles, with its north pole pointing towards Earth's north pole, which is magnetic south.

The value of the angle of the dip here is 90°.

#### Page No 276:

#### Question 10:

Yes, Earth's magnetic field is vertical at the poles. A freely suspended magnet becomes vertical at the poles, with its north pole pointing towards Earth's north pole, which is magnetic south.

The value of the angle of the dip here is 90°.

#### Answer:

(a) Yes, the dip can be zero at the equator of Earth.

(b) Yes, the dip can be 90°â€‹ at the poles of Earth.

#### Page No 276:

#### Question 11:

(a) Yes, the dip can be zero at the equator of Earth.

(b) Yes, the dip can be 90°â€‹ at the poles of Earth.

#### Answer:

Yes, the procedure will work if the instrument is taken to Nepal, as the current at a place can be calculated by multiplying the reduction factor K with tan $\theta $ of that place. In our case, we will take the value of tan $\theta $ of Nepal, as tan $\theta $ may vary from place to place. tan $\theta $ at any place is determined from the mathematical formula $\frac{B}{{B}_{H}}$, where *B* is the external magnetic field and *B _{H}*is the horizontal component of Earth's magnetic field. Thus, we need not take the manual or the instrument back to the factory for correction.

#### Page No 276:

#### Question 1:

Yes, the procedure will work if the instrument is taken to Nepal, as the current at a place can be calculated by multiplying the reduction factor K with tan $\theta $ of that place. In our case, we will take the value of tan $\theta $ of Nepal, as tan $\theta $ may vary from place to place. tan $\theta $ at any place is determined from the mathematical formula $\frac{B}{{B}_{H}}$, where *B* is the external magnetic field and *B _{H}*is the horizontal component of Earth's magnetic field. Thus, we need not take the manual or the instrument back to the factory for correction.

#### Answer:

(a) end-on position

Points lying on the axis of a magnet are called end-on points. In our case, the point on the axis of the loop (on replacing the circular loop with an equivalent magnetic dipole) lies on the axis of the magnetic dipole or on the end-on position.

If P was the point on the axis of the loop, then it is clear from the figure that P lies on the end-on position of the equivalent magnetic dipole.

#### Page No 276:

#### Question 2:

(a) end-on position

Points lying on the axis of a magnet are called end-on points. In our case, the point on the axis of the loop (on replacing the circular loop with an equivalent magnetic dipole) lies on the axis of the magnetic dipole or on the end-on position.

If P was the point on the axis of the loop, then it is clear from the figure that P lies on the end-on position of the equivalent magnetic dipole.

#### Answer:

(b) broadside-on position

The position of the points lying on the equator of a magnetic dipole is called the broadside-on position. In our case, the point on the loop (after replacement of the circular loop with an equivalent magnetic dipole) lies on the equatorial position of the equivalent magnetic dipole. Hence, the point lies on the broadside-on position.

If P was the point on the loop, then it is clear from the figure that point P lies on the broadside-on position of the equivalent magnetic dipole.

#### Page No 276:

#### Question 3:

(b) broadside-on position

The position of the points lying on the equator of a magnetic dipole is called the broadside-on position. In our case, the point on the loop (after replacement of the circular loop with an equivalent magnetic dipole) lies on the equatorial position of the equivalent magnetic dipole. Hence, the point lies on the broadside-on position.

If P was the point on the loop, then it is clear from the figure that point P lies on the broadside-on position of the equivalent magnetic dipole.

#### Answer:

(c) the product *md* is fixed

When we replace a circular current-carrying loop with a magnetic dipole to resemble field lines of the circular loop, the pole strength *m *and the distance between the poles are not fixed.

But the magnetic dipole moment of both systems is always fixed. It is the product of the magnetic moment and the distance between the poles. In other words, *md *is fixed.

A current loop of area *A* and current* I* can be replaced with a magnetic dipole of dipole moment *md.*

i.e*. md *= *IA*

#### Page No 276:

#### Question 4:

(c) the product *md* is fixed

When we replace a circular current-carrying loop with a magnetic dipole to resemble field lines of the circular loop, the pole strength *m *and the distance between the poles are not fixed.

But the magnetic dipole moment of both systems is always fixed. It is the product of the magnetic moment and the distance between the poles. In other words, *md *is fixed.

A current loop of area *A* and current* I* can be replaced with a magnetic dipole of dipole moment *md.*

i.e*. md *= *IA*

#### Answer:

(d) None of these

Magnetic field *B* due to a bar magnet of magnetic moment *M* at distance *r *of the point on the axis* *of the magnet from its centre is given by

$B=\frac{{\mu}_{o}}{4\pi}\frac{2Mr}{{\left({r}^{2}-{l}^{2}\right)}^{2}}$

Here,* 2l *is the length of the magnet.

So, from the above formula, it can be easily seen that $B\propto \frac{r}{{\left({r}^{2}-{l}^{2}\right)}^{2}}$.

#### Page No 276:

#### Question 5:

(d) None of these

Magnetic field *B* due to a bar magnet of magnetic moment *M* at distance *r *of the point on the axis* *of the magnet from its centre is given by

$B=\frac{{\mu}_{o}}{4\pi}\frac{2Mr}{{\left({r}^{2}-{l}^{2}\right)}^{2}}$

Here,* 2l *is the length of the magnet.

So, from the above formula, it can be easily seen that $B\propto \frac{r}{{\left({r}^{2}-{l}^{2}\right)}^{2}}$.

#### Answer:

(c) $\frac{1}{{r}^{3}}$

Magnetic field *B* due to a bar magnet of magnetic moment *M* at distance *r* of the point on the axis* *from its centre is given by

$B=\frac{{\mu}_{\mathit{0}}2Mr}{4\mathrm{\pi}{\left({r}^{2}-{l}^{2}\right)}^{2}}$* 2l *is the length of the magnet.

When the distance of the point where the magnetic field has to be calculated is greater than the length of the magnet, i.e $rl$, the bar magnet acts like a magnetic dipole whose magnetic field is*B*$\propto \frac{1}{{r}^{3}}$

Now,* l* in the denominator can be neglected.

So, the correct option is (c).

#### Page No 276:

#### Question 6:

(c) $\frac{1}{{r}^{3}}$

Magnetic field *B* due to a bar magnet of magnetic moment *M* at distance *r* of the point on the axis* *from its centre is given by

$B=\frac{{\mu}_{\mathit{0}}2Mr}{4\mathrm{\pi}{\left({r}^{2}-{l}^{2}\right)}^{2}}$* 2l *is the length of the magnet.

When the distance of the point where the magnetic field has to be calculated is greater than the length of the magnet, i.e $rl$, the bar magnet acts like a magnetic dipole whose magnetic field is*B*$\propto \frac{1}{{r}^{3}}$

Now,* l* in the denominator can be neglected.

So, the correct option is (c).

#### Answer:

(c) $\frac{{\mu}_{0}}{4\pi}\frac{2\sqrt{2}M}{{d}^{3}}$

Magnetic field (*B*_{1}) due to the short dipole A of dipole moment *M* at an axial point is given by,

${\overrightarrow{B}}_{1}=\frac{{\mu}_{0}}{4\mathrm{\pi}}\frac{2M}{{d}^{3}}...\left(1\right)$

Magnetic field (*B*_{2}) due to the short dipole B of dipole moment *M* at an axial point is given by,

${\overrightarrow{B}}_{2}=\frac{{\mu}_{0}}{4\mathrm{\pi}}\frac{2M}{{d}^{3}}...\left(2\right)$

Resultant magnetic field (*B*) will be,*B* = $\sqrt{{{B}_{1}}^{2}+{{B}_{2}}^{2}}$*B* = $\frac{{\mu}_{0}}{4\mathrm{\pi}}\frac{2\sqrt{2}M}{{d}^{3}}$

#### Page No 276:

#### Question 7:

(c) $\frac{{\mu}_{0}}{4\pi}\frac{2\sqrt{2}M}{{d}^{3}}$

Magnetic field (*B*_{1}) due to the short dipole A of dipole moment *M* at an axial point is given by,

${\overrightarrow{B}}_{1}=\frac{{\mu}_{0}}{4\mathrm{\pi}}\frac{2M}{{d}^{3}}...\left(1\right)$

Magnetic field (*B*_{2}) due to the short dipole B of dipole moment *M* at an axial point is given by,

${\overrightarrow{B}}_{2}=\frac{{\mu}_{0}}{4\mathrm{\pi}}\frac{2M}{{d}^{3}}...\left(2\right)$

Resultant magnetic field (*B*) will be,*B* = $\sqrt{{{B}_{1}}^{2}+{{B}_{2}}^{2}}$*B* = $\frac{{\mu}_{0}}{4\mathrm{\pi}}\frac{2\sqrt{2}M}{{d}^{3}}$

#### Answer:

(d) a vertical plane

Magnetic meridian at a place is not a line but a vertical plane passing through the axis of a freely suspended magnet.

#### Page No 276:

#### Question 8:

(d) a vertical plane

Magnetic meridian at a place is not a line but a vertical plane passing through the axis of a freely suspended magnet.

#### Answer:

(d) will stay in any position

When taken to a geomagnetic pole, a compass needle that is allowed to move in a horizontal plane will try to suspend itself vertically to the horizontal plane containing the compass. In other words, the horizontal plane containing the compass will restrict the compass to suspend itself in vertical direction; hence, the compass will stay in any position.

However, a freely suspended magnet will become vertical at poles, with its north pole pointing towards Earth at its north pole (which is magnetic south).

#### Page No 276:

#### Question 9:

(d) will stay in any position

When taken to a geomagnetic pole, a compass needle that is allowed to move in a horizontal plane will try to suspend itself vertically to the horizontal plane containing the compass. In other words, the horizontal plane containing the compass will restrict the compass to suspend itself in vertical direction; hence, the compass will stay in any position.

However, a freely suspended magnet will become vertical at poles, with its north pole pointing towards Earth at its north pole (which is magnetic south).

#### Answer:

(d) in the direction it is released

At the geomagnetic equator, the needle tries to suspend itself in horizontal direction. But here the needle is restricted to move only in the vertical plane perpendicular to the magnetic meridian. Hence, the needle will stay in the direction it is released.

#### Page No 276:

#### Question 10:

(d) in the direction it is released

At the geomagnetic equator, the needle tries to suspend itself in horizontal direction. But here the needle is restricted to move only in the vertical plane perpendicular to the magnetic meridian. Hence, the needle will stay in the direction it is released.

#### Answer:

(c) curve

Since *i $\propto $*tan $\theta $, the only graph that represents this correlation is curve *c*.

#### Page No 276:

#### Question 11:

(c) curve

Since *i $\propto $*tan $\theta $, the only graph that represents this correlation is curve *c*.

#### Answer:

(c) remain unchanged

For a tangent galvanometer, deflection is given by

$\theta ={\mathrm{tan}}^{-1}\left(\frac{i}{k}\right)$

Here, *k* is the constant called reduction factor.

From the above formula, we can say that deflection is independent of the number of turns.

Hence, on doubling the number of turns, deflection remains the same.

#### Page No 276:

#### Question 12:

(c) remain unchanged

For a tangent galvanometer, deflection is given by

$\theta ={\mathrm{tan}}^{-1}\left(\frac{i}{k}\right)$

Here, *k* is the constant called reduction factor.

From the above formula, we can say that deflection is independent of the number of turns.

Hence, on doubling the number of turns, deflection remains the same.

#### Answer:

(b) a moving coil galvanometer

The current and deflection dependence of a moving coil galvanometer is given by

$i=\frac{k}{nAB}\theta \Rightarrow i\propto \theta $

Therefore, if we double the current, the deflection also gets doubled.

However, in a tangent galvanometer, $i\propto \mathrm{tan}\theta $; that is, there is no direct relation between $\theta $ and current.

Hence, the correct option is (b).

#### Page No 276:

#### Question 13:

(b) a moving coil galvanometer

The current and deflection dependence of a moving coil galvanometer is given by

$i=\frac{k}{nAB}\theta \Rightarrow i\propto \theta $

Therefore, if we double the current, the deflection also gets doubled.

However, in a tangent galvanometer, $i\propto \mathrm{tan}\theta $; that is, there is no direct relation between $\theta $ and current.

Hence, the correct option is (b).

#### Answer:

(a) very nearly 2*$\mathrm{\pi}$ai*B perpendicular to the plane of the wire

In this case, the north pole of the magnet is coinciding with the centre of the circular loop carrying electric current *i.* So, the magnetic field lines almost lie on the plane of the ring and the force due to the field lines is perpendicular to the field lines and to the plane of the circular ring.

Let *idl *be the current element, B be the magnetic field and *dF *be the force on the current element* idl.*

Now*dF = *B*idl* $\Rightarrow F={\int}_{0}^{2\mathrm{\pi a}}\mathrm{B}idl$

$\Rightarrow F=2\mathrm{\pi a}i\mathrm{B}$

Thus, the force acting on the wire is 2$\mathrm{\pi}$a*i*B and it is perpendicular to the plane of the wire.

#### Page No 277:

#### Question 1:

(a) very nearly 2*$\mathrm{\pi}$ai*B perpendicular to the plane of the wire

In this case, the north pole of the magnet is coinciding with the centre of the circular loop carrying electric current *i.* So, the magnetic field lines almost lie on the plane of the ring and the force due to the field lines is perpendicular to the field lines and to the plane of the circular ring.

Let *idl *be the current element, B be the magnetic field and *dF *be the force on the current element* idl.*

Now*dF = *B*idl* $\Rightarrow F={\int}_{0}^{2\mathrm{\pi a}}\mathrm{B}idl$

$\Rightarrow F=2\mathrm{\pi a}i\mathrm{B}$

Thus, the force acting on the wire is 2$\mathrm{\pi}$a*i*B and it is perpendicular to the plane of the wire.

#### Answer:

(a) Magnetic field is produced by electric charges only.

(b) Magnetic poles are only mathematical assumptions having no real existence.

Justification of (a) and (b):

Investigators and experimenters have failed to find any sign of magnetic monopoles. So, we can assume that magnetic monopoles are only a mathematical assumption.

A magnetic field is produced by the motion of an electric charge only. In paramagnets or ferromagnets, the motion of an electron (charge) and the alignment of domains (bunch of charges with particular alignment) create paramagnetism and ferromagnetism, respectively.

Therefore, the only cause behind the magnetic field is the motion of an electric charge.

Denial of (c):

The north pole is equivalent to an anticlockwise current and the south pole is equivalent to a clockwise current.

Denial of (d):

A bar magnet is not equivalent to a long, straight current because the distribution and orientation of magnetic field lines do not resemble each other.

#### Page No 277:

#### Question 2:

(a) Magnetic field is produced by electric charges only.

(b) Magnetic poles are only mathematical assumptions having no real existence.

Justification of (a) and (b):

Investigators and experimenters have failed to find any sign of magnetic monopoles. So, we can assume that magnetic monopoles are only a mathematical assumption.

A magnetic field is produced by the motion of an electric charge only. In paramagnets or ferromagnets, the motion of an electron (charge) and the alignment of domains (bunch of charges with particular alignment) create paramagnetism and ferromagnetism, respectively.

Therefore, the only cause behind the magnetic field is the motion of an electric charge.

Denial of (c):

The north pole is equivalent to an anticlockwise current and the south pole is equivalent to a clockwise current.

Denial of (d):

A bar magnet is not equivalent to a long, straight current because the distribution and orientation of magnetic field lines do not resemble each other.

#### Answer:

(b) The line SN should be perpendicular to the plane of the loop

(d) The north pole should be below the loop.

A horizontal circular loop carrying current in clockwise direction acts like the south pole of a magnet. Hence, the south pole of the magnet coincides with the loop.

Now, when the loop carrying current in clockwise direction is viewed from above, it looks like the magnetic lines of force are entering the loop thus it acts like south pole of a magnet. And if we view from below the loop then it appears that magnetic lines of force are leaving the loop. Hence, the north pole should be below the loop.

#### Page No 277:

#### Question 3:

(b) The line SN should be perpendicular to the plane of the loop

(d) The north pole should be below the loop.

A horizontal circular loop carrying current in clockwise direction acts like the south pole of a magnet. Hence, the south pole of the magnet coincides with the loop.

Now, when the loop carrying current in clockwise direction is viewed from above, it looks like the magnetic lines of force are entering the loop thus it acts like south pole of a magnet. And if we view from below the loop then it appears that magnetic lines of force are leaving the loop. Hence, the north pole should be below the loop.

#### Answer:

(a) P_{1} and P_{2}

(b) Q_{1} and Q_{2}

We know that magnetic field lines are directed from the north pole to the south pole. From the given figure, we can say that the direction of magnetic field $\underset{B}{\to}$ is the same only at points P_{1} and P_{2} and at points Q_{1} and Q_{2}.

#### Page No 277:

#### Question 4:

(a) P_{1} and P_{2}

(b) Q_{1} and Q_{2}

We know that magnetic field lines are directed from the north pole to the south pole. From the given figure, we can say that the direction of magnetic field $\underset{B}{\to}$ is the same only at points P_{1} and P_{2} and at points Q_{1} and Q_{2}.

#### Answer:

(c) P_{1} and Q_{1}

(d) P_{2} and Q_{2}

We know that magnetic field lines are directed from the north pole to the south pole. From the given figure, we can say that the direction of the magnetic field $\underset{B}{\to}$ is opposite at points P_{1} and Q_{1} and at points P_{2} and Q_{2}.

#### Page No 277:

#### Question 5:

(c) P_{1} and Q_{1}

(d) P_{2} and Q_{2}

We know that magnetic field lines are directed from the north pole to the south pole. From the given figure, we can say that the direction of the magnetic field $\underset{B}{\to}$ is opposite at points P_{1} and Q_{1} and at points P_{2} and Q_{2}.

#### Answer:

(b) a deflection galvanometer if the earth's horizontal field is known

(c) an oscillation magnetometer if the earth's horizontal field is known

(d) both deflection and oscillation magnetometers if the earth's horizontal field is not known

Denial of (a):

Tangent galvanometer is an instrument used to measure electric current; it cannot be used to the measure magnetic moment of a bar magnet.

Justification of (b) and (c):

Deflection magnetometer is used to measure $\frac{M}{{B}_{H}}$ of a permanent bar magnet.

Similarly, oscillation magnetometer is used to measure *M B _{H}*of a bar magnet. So, if earth's horizontal field,

*B*, is known, then the magnetic moment of a bar magnet,

_{H}*M,*can be measured.

Justification of (d):

Using deflection and oscillation magnetometers, we can calculate $\frac{M}{{B}_{H}}$ and

*M B*, respectively. Therefore, if we multiply the result obtained from both the instruments, then

_{H}*B*cancels out as $\frac{M}{{B}_{H}}$$\times $

_{H}*M B*â€‹ =

_{H}*M*Thus, the value of

^{2}.*B*is not required.

_{H}Therefore, we can use both deflection and oscillation magnetometers if the earth's horizontal field is not known.

#### Page No 277:

#### Question 1:

(b) a deflection galvanometer if the earth's horizontal field is known

(c) an oscillation magnetometer if the earth's horizontal field is known

(d) both deflection and oscillation magnetometers if the earth's horizontal field is not known

Denial of (a):

Tangent galvanometer is an instrument used to measure electric current; it cannot be used to the measure magnetic moment of a bar magnet.

Justification of (b) and (c):

Deflection magnetometer is used to measure $\frac{M}{{B}_{H}}$ of a permanent bar magnet.

Similarly, oscillation magnetometer is used to measure *M B _{H}*of a bar magnet. So, if earth's horizontal field,

*B*, is known, then the magnetic moment of a bar magnet,

_{H}*M,*can be measured.

Justification of (d):

Using deflection and oscillation magnetometers, we can calculate $\frac{M}{{B}_{H}}$ and

*M B*, respectively. Therefore, if we multiply the result obtained from both the instruments, then

_{H}*B*cancels out as $\frac{M}{{B}_{H}}$$\times $

_{H}*M B*â€‹ =

_{H}*M*Thus, the value of

^{2}.*B*is not required.

_{H}Therefore, we can use both deflection and oscillation magnetometers if the earth's horizontal field is not known.

#### Answer:

Given:

Pole strength of the bar magnet, *m* = 10 Am

Distance of the point from the north pole of the bar magnet, *r *= 5 cm = 0.05 m

We know,

The magnetic field due to magnetic charge $\left(B\right)$ is given by

$B=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{m}{{r}^{\mathit{2}}}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{-7}\times 10}{{\left(5\times {10}^{-2}\right)}^{2}}=\frac{{10}^{-6}}{25\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{-2}}{25}=4\times {10}^{-4}\mathrm{T}$

#### Page No 277:

#### Question 2:

Given:

Pole strength of the bar magnet, *m* = 10 Am

Distance of the point from the north pole of the bar magnet, *r *= 5 cm = 0.05 m

We know,

The magnetic field due to magnetic charge $\left(B\right)$ is given by

$B=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{m}{{r}^{\mathit{2}}}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{-7}\times 10}{{\left(5\times {10}^{-2}\right)}^{2}}=\frac{{10}^{-6}}{25\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{-2}}{25}=4\times {10}^{-4}\mathrm{T}$

#### Answer:

Given:

Pole strength = *m*_{1}_{}= *m*_{2} = 10 Am

Distance between the north pole of the first magnet and the south pole of the second magnet, *r* = 2 cm = 0.02 m

We know,

Force $\left(F\right)$ exerted by two magnetic poles on each other is given by

$F=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{{m}_{1}{m}_{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\mathrm{\pi}\times {10}^{-7}\times {10}^{2}}{4\mathrm{\pi}\times 4\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}=2.5\times {10}^{-2}\mathrm{N}$

#### Page No 277:

#### Question 3:

Given:

Pole strength = *m*_{1}_{}= *m*_{2} = 10 Am

Distance between the north pole of the first magnet and the south pole of the second magnet, *r* = 2 cm = 0.02 m

We know,

Force $\left(F\right)$ exerted by two magnetic poles on each other is given by

$F=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{{m}_{1}{m}_{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\mathrm{\pi}\times {10}^{-7}\times {10}^{2}}{4\mathrm{\pi}\times 4\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}=2.5\times {10}^{-2}\mathrm{N}$

#### Answer:

Given:

Magnetic field in the space, *B* = 0.20 × 10^{−3} T

Distance moved, $\u2206r$ = 50 cm

We know,

$B=-\frac{dV}{dl}$

$\Rightarrow dV\mathit{}\mathit{=}\mathit{}{\mathit{\int}}_{{r}_{1}}^{{r}_{2}}\mathit{}\mathit{-}\stackrel{\mathit{\rightharpoonup}}{B\mathit{.}}\mathrm{d}l\phantom{\rule{0ex}{0ex}}$

Since the magnetic field is uniform, it can come out of the integration sign.

$\Rightarrow \u2206V\mathit{}\mathit{=}\mathit{}\mathit{-}B\mathit{.}(\u2206r)\phantom{\rule{0ex}{0ex}}$

Here, $\u2206V$ is the change in the potential.

∴ Change in the potential = −0.2 × 10^{−3} × 0.5

= −0.1 × 10^{−3} T-m

Here, the negative sign shows that the potential decreases.

#### Page No 277:

#### Question 4:

Given:

Magnetic field in the space, *B* = 0.20 × 10^{−3} T

Distance moved, $\u2206r$ = 50 cm

We know,

$B=-\frac{dV}{dl}$

$\Rightarrow dV\mathit{}\mathit{=}\mathit{}{\mathit{\int}}_{{r}_{1}}^{{r}_{2}}\mathit{}\mathit{-}\stackrel{\mathit{\rightharpoonup}}{B\mathit{.}}\mathrm{d}l\phantom{\rule{0ex}{0ex}}$

Since the magnetic field is uniform, it can come out of the integration sign.

$\Rightarrow \u2206V\mathit{}\mathit{=}\mathit{}\mathit{-}B\mathit{.}(\u2206r)\phantom{\rule{0ex}{0ex}}$

Here, $\u2206V$ is the change in the potential.

∴ Change in the potential = −0.2 × 10^{−3} × 0.5

= −0.1 × 10^{−3} T-m

Here, the negative sign shows that the potential decreases.

#### Answer:

Given:

Perpendicular distance, *dx* = 10 sin30^{â‚’} cm = 0.05 m,

Change in the potential, *dV* = 0.1 × 10^{−4} T-m

We know that the relation between the potential and the field is given by

$B=-\frac{dV}{dx}\phantom{\rule{0ex}{0ex}}\Rightarrow B=-\frac{0.1\times {10}^{-4}\mathrm{T}-m}{5\times {10}^{-2}m}\phantom{\rule{0ex}{0ex}}\Rightarrow B=-2\times {10}^{-4}\mathrm{T}$*B* is perpendicular to the equipotential surface. Here, it is at angle of 120° with the positive *x*-axis.

#### Page No 277:

#### Question 5:

Given:

Perpendicular distance, *dx* = 10 sin30^{â‚’} cm = 0.05 m,

Change in the potential, *dV* = 0.1 × 10^{−4} T-m

We know that the relation between the potential and the field is given by

$B=-\frac{dV}{dx}\phantom{\rule{0ex}{0ex}}\Rightarrow B=-\frac{0.1\times {10}^{-4}\mathrm{T}-m}{5\times {10}^{-2}m}\phantom{\rule{0ex}{0ex}}\Rightarrow B=-2\times {10}^{-4}\mathrm{T}$*B* is perpendicular to the equipotential surface. Here, it is at angle of 120° with the positive *x*-axis.

#### Answer:

Given:

Magnetic field strength, *B* = 2 × 10^{−4} T

Distance of the point from the dipole*, d* = 10 cm = 0.1 m

(a) If the point is at the end-on position:

The magnetic field $\left(B\right)$ on the axial point of the dipole is given by

$B=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{2M}{{d}^{3}}\phantom{\rule{0ex}{0ex}}\mathrm{Here},M\mathrm{is}\mathrm{the}\mathrm{magnetic}\mathrm{moment}\mathrm{of}\mathrm{the}\mathrm{dipole}\mathrm{that}\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{find}\mathrm{out}.\phantom{\rule{0ex}{0ex}}\therefore 2\times {10}^{-4}=\frac{{10}^{-7}\times 2M}{{\left({10}^{-1}\right)}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow M=\frac{2\times {10}^{-4}\times {10}^{-3}}{{10}^{-7}\times 2}\phantom{\rule{0ex}{0ex}}\Rightarrow M=1\mathrm{A}-{\mathrm{m}}^{2}$

(b) If the point is at broadside-on position (equatorial position):

The magnetic field $\left(B\right)$ is given by

$B=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{\mathit{M}}{{\mathit{d}}^{\mathit{3}}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2\times {10}^{-4}=\frac{{10}^{-7}\times M}{{\left({10}^{-1}\right)}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow M=2\mathrm{A}-{\mathrm{m}}^{2}$

#### Page No 277:

#### Question 6:

Given:

Magnetic field strength, *B* = 2 × 10^{−4} T

Distance of the point from the dipole*, d* = 10 cm = 0.1 m

(a) If the point is at the end-on position:

The magnetic field $\left(B\right)$ on the axial point of the dipole is given by

$B=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{2M}{{d}^{3}}\phantom{\rule{0ex}{0ex}}\mathrm{Here},M\mathrm{is}\mathrm{the}\mathrm{magnetic}\mathrm{moment}\mathrm{of}\mathrm{the}\mathrm{dipole}\mathrm{that}\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{find}\mathrm{out}.\phantom{\rule{0ex}{0ex}}\therefore 2\times {10}^{-4}=\frac{{10}^{-7}\times 2M}{{\left({10}^{-1}\right)}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow M=\frac{2\times {10}^{-4}\times {10}^{-3}}{{10}^{-7}\times 2}\phantom{\rule{0ex}{0ex}}\Rightarrow M=1\mathrm{A}-{\mathrm{m}}^{2}$

(b) If the point is at broadside-on position (equatorial position):

The magnetic field $\left(B\right)$ is given by

$B=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{\mathit{M}}{{\mathit{d}}^{\mathit{3}}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2\times {10}^{-4}=\frac{{10}^{-7}\times M}{{\left({10}^{-1}\right)}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow M=2\mathrm{A}-{\mathrm{m}}^{2}$

#### Answer:

Given:

Angle made by observation point P with the axis of the dipole, $\theta $ = ${\mathrm{tan}}^{-1}\left(\sqrt{2}\right)$

$\Rightarrow \mathrm{tan}\theta =\sqrt{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2={\mathrm{tan}}^{2}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\mathrm{cot}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{tan}\theta}{2}=\mathrm{cot}\theta ....\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know},\phantom{\rule{0ex}{0ex}}\frac{\mathrm{tan}\theta}{2}=\mathrm{tan}\alpha ....\left(2\right)$

On comparing (1) and (2), we get

tan α = cot *θ*

$\Rightarrow $tan α = tan (90 − *θ*)

$\Rightarrow $α = 90 − *θ*

$\Rightarrow $*θ* + α = 90°

Hence, the magnetic field due to the dipole is perpendicular to the magnetic axis.

#### Page No 277:

#### Question 7:

Given:

Angle made by observation point P with the axis of the dipole, $\theta $ = ${\mathrm{tan}}^{-1}\left(\sqrt{2}\right)$

$\Rightarrow \mathrm{tan}\theta =\sqrt{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2={\mathrm{tan}}^{2}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\mathrm{cot}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{tan}\theta}{2}=\mathrm{cot}\theta ....\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know},\phantom{\rule{0ex}{0ex}}\frac{\mathrm{tan}\theta}{2}=\mathrm{tan}\alpha ....\left(2\right)$

On comparing (1) and (2), we get

tan α = cot *θ*

$\Rightarrow $tan α = tan (90 − *θ*)

$\Rightarrow $α = 90 − *θ*

$\Rightarrow $*θ* + α = 90°

Hence, the magnetic field due to the dipole is perpendicular to the magnetic axis.

#### Answer:

Given:

Length of the magnet, 2*l* = 8 cm = 8 $\times $ 10^{$-2$} m

Distance of the observation point from the centre of the dipole, *d* = 3 cm

Magnetic field in the broadside-on position, *B* = 4 × 10^{−6} T

The magnetic field due to the dipole on the equatorial point $\left(B\right)$is given by

$B=\frac{{\mathrm{\mu}}_{0}m2l}{4\mathrm{\pi}{\left({d}^{2}+{l}^{2}\right)}^{3/2}}\phantom{\rule{0ex}{0ex}}$

Here, *m* is the pole strength of the magnet.

On substituting the respective values, we get

$4\times {10}^{-6}=\frac{{10}^{-7}m\times 8\times {10}^{-2}}{{\left(9\times {10}^{-4}+16\times {10}^{-4}\right)}^{3/2}}\phantom{\rule{0ex}{0ex}}\Rightarrow 4\times {10}^{-6}=\frac{m\times 8\times {10}^{-9}}{{\left(25\right)}^{3/2}\times {\left({10}^{-4}\right)}^{3/2}}\phantom{\rule{0ex}{0ex}}\Rightarrow m=\frac{4\times {10}^{-6}\times 125\times {10}^{-6}}{8\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}\Rightarrow m=6.25\times {10}^{-2}\mathrm{A}-\mathrm{m}$

Thus, the pole strength of the magnet is 6.25$\times $10^{$-2$}A-m.

#### Page No 277:

#### Question 8:

Given:

Length of the magnet, 2*l* = 8 cm = 8 $\times $ 10^{$-2$} m

Distance of the observation point from the centre of the dipole, *d* = 3 cm

Magnetic field in the broadside-on position, *B* = 4 × 10^{−6} T

The magnetic field due to the dipole on the equatorial point $\left(B\right)$is given by

$B=\frac{{\mathrm{\mu}}_{0}m2l}{4\mathrm{\pi}{\left({d}^{2}+{l}^{2}\right)}^{3/2}}\phantom{\rule{0ex}{0ex}}$

Here, *m* is the pole strength of the magnet.

On substituting the respective values, we get

$4\times {10}^{-6}=\frac{{10}^{-7}m\times 8\times {10}^{-2}}{{\left(9\times {10}^{-4}+16\times {10}^{-4}\right)}^{3/2}}\phantom{\rule{0ex}{0ex}}\Rightarrow 4\times {10}^{-6}=\frac{m\times 8\times {10}^{-9}}{{\left(25\right)}^{3/2}\times {\left({10}^{-4}\right)}^{3/2}}\phantom{\rule{0ex}{0ex}}\Rightarrow m=\frac{4\times {10}^{-6}\times 125\times {10}^{-6}}{8\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}\Rightarrow m=6.25\times {10}^{-2}\mathrm{A}-\mathrm{m}$

Thus, the pole strength of the magnet is 6.25$\times $10^{$-2$}A-m.

#### Answer:

Given:

Magnetic moment of the magnetic dipole, *M = *1.44 A-m^{2}

â€‹Horizontal component of Earth's magnetic field, *B*_{H}_{ }= 18 μT

We know that for a magnetic dipole with pole pointing to the north, the neutral point always lies in the broadside-on position.

Let *d* be the perpendicular distance of the neutral point from mid point of the magnet.

The magnetic field due to the dipole at the broadside-on position (*B*) is given by

$\overrightarrow{B}\mathit{}=\frac{{\mathrm{\mu}}_{0}M}{4\mathrm{\pi}{d}^{3}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

This magnetic field strength should be equal to the horizontal component of Earth's magnetic field at that point, that is, *B*_{H}_{ }due to Earth.

Thus,

$\overrightarrow{B}\mathit{}=\frac{{\mathrm{\mu}}_{0}M}{4\mathrm{\pi}{d}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{10}^{-7}\times 1.44}{{d}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow 18\times {10}^{-6}=\frac{{10}^{-7}\times 1.44}{{d}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{\mathit{3}}=8\times {10}^{-3}\phantom{\rule{0ex}{0ex}}\Rightarrow d=2\times {10}^{-1}=20\mathrm{cm}$

#### Page No 277:

#### Question 9:

Given:

Magnetic moment of the magnetic dipole, *M = *1.44 A-m^{2}

â€‹Horizontal component of Earth's magnetic field, *B*_{H}_{ }= 18 μT

We know that for a magnetic dipole with pole pointing to the north, the neutral point always lies in the broadside-on position.

Let *d* be the perpendicular distance of the neutral point from mid point of the magnet.

The magnetic field due to the dipole at the broadside-on position (*B*) is given by

$\overrightarrow{B}\mathit{}=\frac{{\mathrm{\mu}}_{0}M}{4\mathrm{\pi}{d}^{3}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

This magnetic field strength should be equal to the horizontal component of Earth's magnetic field at that point, that is, *B*_{H}_{ }due to Earth.

Thus,

$\overrightarrow{B}\mathit{}=\frac{{\mathrm{\mu}}_{0}M}{4\mathrm{\pi}{d}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{10}^{-7}\times 1.44}{{d}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow 18\times {10}^{-6}=\frac{{10}^{-7}\times 1.44}{{d}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{\mathit{3}}=8\times {10}^{-3}\phantom{\rule{0ex}{0ex}}\Rightarrow d=2\times {10}^{-1}=20\mathrm{cm}$

#### Answer:

Given:

Magnetic moment of the magnetic dipole, *M = *0.72 Am^{2}

â€‹Horizontal component of Earth's magnetic field, *B*_{H}â€‹ = 18 μT

â€‹Let *d* be the distance of the neutral point from the south of the dipole.

When the magnet is such that its north pole faces the geographic south of Earth, the neutral point lies along the axial line of the magnet.

Thus, the magnetic field on the axial point of the dipole (*B*) is given by

$B=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{2M}{{d}^{3}}\phantom{\rule{0ex}{0ex}}$

This magnetic field strength should be equal to the horizontal component of Earth's magnetic field.

Thus,

$\frac{{10}^{-7}\times 2\times 0.72}{{d}^{3}}=18\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{3}=\frac{2\times 0.72\times {10}^{-7}}{18\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}\Rightarrow d={\left(\frac{8\times {10}^{-9}}{{10}^{-6}}\right)}^{1/3}\phantom{\rule{0ex}{0ex}}\Rightarrow d=2\times {10}^{-1}\mathrm{m}=20\mathrm{cm}$

#### Page No 277:

#### Question 10:

Given:

Magnetic moment of the magnetic dipole, *M = *0.72 Am^{2}

â€‹Horizontal component of Earth's magnetic field, *B*_{H}â€‹ = 18 μT

â€‹Let *d* be the distance of the neutral point from the south of the dipole.

When the magnet is such that its north pole faces the geographic south of Earth, the neutral point lies along the axial line of the magnet.

Thus, the magnetic field on the axial point of the dipole (*B*) is given by

$B=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{2M}{{d}^{3}}\phantom{\rule{0ex}{0ex}}$

This magnetic field strength should be equal to the horizontal component of Earth's magnetic field.

Thus,

$\frac{{10}^{-7}\times 2\times 0.72}{{d}^{3}}=18\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{3}=\frac{2\times 0.72\times {10}^{-7}}{18\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}\Rightarrow d={\left(\frac{8\times {10}^{-9}}{{10}^{-6}}\right)}^{1/3}\phantom{\rule{0ex}{0ex}}\Rightarrow d=2\times {10}^{-1}\mathrm{m}=20\mathrm{cm}$

#### Answer:

Given:

Magnetic moment of magnetic dipole, *M* $=0.72\sqrt{2}\mathrm{A}-{\mathrm{m}}^{2}$

Horizontal component of the earth's magnetic field,* **B*_{H} = 18 μT

Let *d* be the distance of neutral point from the dipole.

Magnetic field due to the bar magnet $\left(B\right)$ on the equatorial line of the dipole is given by,

$B=\frac{{\mu}_{0}}{4\pi}\frac{M}{{d}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4\mathrm{\pi}\times {10}^{-7}}{4\mathrm{\pi}}\times \frac{0.72\sqrt{2}}{{d}^{3}}=18\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{3}=\frac{0.72\times 1.414\times {10}^{-7}}{18\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{3}=0.005656\phantom{\rule{0ex}{0ex}}\Rightarrow d=\sqrt[3]{0.005656}\phantom{\rule{0ex}{0ex}}\Rightarrow d\approx 0.2\mathrm{m}=20\mathrm{cm}$

#### Page No 277:

#### Question 11:

Given:

Magnetic moment of magnetic dipole, *M* $=0.72\sqrt{2}\mathrm{A}-{\mathrm{m}}^{2}$

Horizontal component of the earth's magnetic field,* **B*_{H} = 18 μT

Let *d* be the distance of neutral point from the dipole.

Magnetic field due to the bar magnet $\left(B\right)$ on the equatorial line of the dipole is given by,

$B=\frac{{\mu}_{0}}{4\pi}\frac{M}{{d}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4\mathrm{\pi}\times {10}^{-7}}{4\mathrm{\pi}}\times \frac{0.72\sqrt{2}}{{d}^{3}}=18\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{3}=\frac{0.72\times 1.414\times {10}^{-7}}{18\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{3}=0.005656\phantom{\rule{0ex}{0ex}}\Rightarrow d=\sqrt[3]{0.005656}\phantom{\rule{0ex}{0ex}}\Rightarrow d\approx 0.2\mathrm{m}=20\mathrm{cm}$

#### Answer:

Given:

Magnetic moment of dipole at Earth's centre, *M* = 8.0 × 10^{22} Am^{2}

Radius of Earth, *d* = 6400 km

The geomagnetic pole is at the end of the position (axial) of Earth.

Thus, the magnetic field at the axial point of the dipole $\left(B\right)$ is given by

$B=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{\mathit{2}M}{{d}^{\mathit{3}}}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{10}^{-7}\times 2\times 8\times {10}^{22}}{{\left(64\right)}^{3}\times {10}^{15}}\phantom{\rule{0ex}{0ex}}\Rightarrow B=6\times {10}^{-5}\mathrm{T}=60\mathrm{\mu T}$

#### Page No 277:

#### Question 12:

Given:

Magnetic moment of dipole at Earth's centre, *M* = 8.0 × 10^{22} Am^{2}

Radius of Earth, *d* = 6400 km

The geomagnetic pole is at the end of the position (axial) of Earth.

Thus, the magnetic field at the axial point of the dipole $\left(B\right)$ is given by

$B=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{\mathit{2}M}{{d}^{\mathit{3}}}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{10}^{-7}\times 2\times 8\times {10}^{22}}{{\left(64\right)}^{3}\times {10}^{15}}\phantom{\rule{0ex}{0ex}}\Rightarrow B=6\times {10}^{-5}\mathrm{T}=60\mathrm{\mu T}$

#### Answer:

Given:

Magnetic field at the magnetic equator, *B = *3.4 × 10^{−5} T

Let *M* be the magnetic moment of Earth's magnetic dipole and R be the distance of the observation point from the centre of Earth's magnetic dipole.

As the point on the magnetic equator is on the equatorial position of Earth's magnet, the magnetic field at the equatorial point $\left(B\right)$ is given by

$B=\frac{{\mu}_{0}}{4\mathrm{\pi}}\frac{M}{{R}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\mu}_{0}}{4\mathrm{\pi}}\times \frac{M}{{R}^{\mathit{3}}}=3.4\times {10}^{-5}\phantom{\rule{0ex}{0ex}}\Rightarrow M=\frac{3.4\times {10}^{-5}\times {\mathrm{R}}^{3}\times 4\mathrm{\pi}}{4\mathrm{\pi}\times {10}^{-7}}\phantom{\rule{0ex}{0ex}}\Rightarrow M=3.4\times {10}^{2}{R}^{3}\phantom{\rule{0ex}{0ex}}$

As the magnetic field on Earth's geomagnetic poles lies on the axial point of the magnetic dipole, the magnetic field at the axial point $\left({B}_{1}\right)$ is given by

${\overrightarrow{B}}_{1}=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\times \frac{2M}{{R}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{1}=\frac{{10}^{-7}\times 2\times 3.4\times {10}^{2}\times {R}^{3}}{{R}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{1}=6.8\times {10}^{-5}\mathrm{T}$

#### Page No 277:

#### Question 13:

Given:

Magnetic field at the magnetic equator, *B = *3.4 × 10^{−5} T

Let *M* be the magnetic moment of Earth's magnetic dipole and R be the distance of the observation point from the centre of Earth's magnetic dipole.

As the point on the magnetic equator is on the equatorial position of Earth's magnet, the magnetic field at the equatorial point $\left(B\right)$ is given by

$B=\frac{{\mu}_{0}}{4\mathrm{\pi}}\frac{M}{{R}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\mu}_{0}}{4\mathrm{\pi}}\times \frac{M}{{R}^{\mathit{3}}}=3.4\times {10}^{-5}\phantom{\rule{0ex}{0ex}}\Rightarrow M=\frac{3.4\times {10}^{-5}\times {\mathrm{R}}^{3}\times 4\mathrm{\pi}}{4\mathrm{\pi}\times {10}^{-7}}\phantom{\rule{0ex}{0ex}}\Rightarrow M=3.4\times {10}^{2}{R}^{3}\phantom{\rule{0ex}{0ex}}$

As the magnetic field on Earth's geomagnetic poles lies on the axial point of the magnetic dipole, the magnetic field at the axial point $\left({B}_{1}\right)$ is given by

${\overrightarrow{B}}_{1}=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\times \frac{2M}{{R}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{1}=\frac{{10}^{-7}\times 2\times 3.4\times {10}^{2}\times {R}^{3}}{{R}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{1}=6.8\times {10}^{-5}\mathrm{T}$

#### Answer:

Given:

Horizontal component of Earth's magnetic field, *B = *26 μT â€‹

Angle of dip, δ = 60°

The horizontal component of Earth's magnetic field $\left({B}_{H}\right)$ is given by*B _{H}* =

*B*cosδ

Here,

*B*= Total magnetic field of Earth

On substituting the respective values, we get

$26\times {10}^{-6}=B\times \frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow B=52\times {10}^{-6}=52\mathrm{\mu T}\phantom{\rule{0ex}{0ex}}$

The vertical component of Earth's magnetic field $\left({B}_{y}\right)$ is given by

${B}_{y}\mathit{=}B\mathrm{sin}\delta \phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{y}=52\times {10}^{-6}\times \frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{y}=44.98\mathrm{\mu T}=45\mathrm{\mu T}$

#### Page No 277:

#### Question 14:

Given:

Horizontal component of Earth's magnetic field, *B = *26 μT â€‹

Angle of dip, δ = 60°

The horizontal component of Earth's magnetic field $\left({B}_{H}\right)$ is given by*B _{H}* =

*B*cosδ

Here,

*B*= Total magnetic field of Earth

On substituting the respective values, we get

$26\times {10}^{-6}=B\times \frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow B=52\times {10}^{-6}=52\mathrm{\mu T}\phantom{\rule{0ex}{0ex}}$

The vertical component of Earth's magnetic field $\left({B}_{y}\right)$ is given by

${B}_{y}\mathit{=}B\mathrm{sin}\delta \phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{y}=52\times {10}^{-6}\times \frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{y}=44.98\mathrm{\mu T}=45\mathrm{\mu T}$

#### Answer:

Given:

Angle made by the magnetic meridian with the plane of rotation of the needle, $\theta =60\xb0$

Angle made by the needle with the horizontal, ${\delta}_{1}={\mathrm{tan}}^{-1}\left(\frac{2}{\sqrt{3}}\right)$

If $\delta $ is the angle of dip, then

$\mathrm{tan}{\delta}_{1}=\frac{\mathrm{tan}\delta}{\mathrm{cos}\theta}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\delta =\mathrm{tan}{\delta}_{1}\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\delta =\mathrm{tan}\left({\mathrm{tan}}^{-1}\frac{2}{\sqrt{3}}\right)\mathrm{cos}60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\delta =\frac{2}{\sqrt{3}}\times \frac{1}{2}=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \delta =30\xb0$

#### Page No 278:

#### Question 15:

Given:

Angle made by the magnetic meridian with the plane of rotation of the needle, $\theta =60\xb0$

Angle made by the needle with the horizontal, ${\delta}_{1}={\mathrm{tan}}^{-1}\left(\frac{2}{\sqrt{3}}\right)$

If $\delta $ is the angle of dip, then

$\mathrm{tan}{\delta}_{1}=\frac{\mathrm{tan}\delta}{\mathrm{cos}\theta}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\delta =\mathrm{tan}{\delta}_{1}\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\delta =\mathrm{tan}\left({\mathrm{tan}}^{-1}\frac{2}{\sqrt{3}}\right)\mathrm{cos}60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\delta =\frac{2}{\sqrt{3}}\times \frac{1}{2}=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \delta =30\xb0$

#### Answer:

Given:

Apparent dip shown by the needle of the dip circle, *δ*_{1} = 45°â€‹

Dip shown by the needle of the dip circle on rotating the circle, â€‹*δ*_{2 }= 53° â€‹

True dip $\left(\delta \right)$ is given by

Cot^{2}* δ* = Cot^{2}* δ*_{1} + Cot^{2}* δ*_{2}

$\Rightarrow $Cot^{2 }*δ* = Cot^{2} 45° + Cot^{2} 53°

$\Rightarrow $Cot^{2 }*δ* = 1.56

$\Rightarrow $Cot^{2}^{}*δ* = 1.56

$\Rightarrow $*δ* = 38.6° ≈ 39°

#### Page No 278:

#### Question 16:

Given:

Apparent dip shown by the needle of the dip circle, *δ*_{1} = 45°â€‹

Dip shown by the needle of the dip circle on rotating the circle, â€‹*δ*_{2 }= 53° â€‹

True dip $\left(\delta \right)$ is given by

Cot^{2}* δ* = Cot^{2}* δ*_{1} + Cot^{2}* δ*_{2}

$\Rightarrow $Cot^{2 }*δ* = Cot^{2} 45° + Cot^{2} 53°

$\Rightarrow $Cot^{2 }*δ* = 1.56

$\Rightarrow $Cot^{2}^{}*δ* = 1.56

$\Rightarrow $*δ* = 38.6° ≈ 39°

#### Answer:

Given:

Horizontal component of Earth's magnetic field, *B*_{H} = 3.6 × 10^{−5} T

Deflection shown by the tangent galvanometer,* θ *= 45°

Current through the galvanometer, *I* = 10 mA = 10^{−2} A

Radius of the coil, *r *= 10 cm = 0.1 m

Number of turns in the coil, *n* = ?

We know,

${B}_{\mathrm{H}}\mathrm{tan}\theta =\frac{{\mathrm{\mu}}_{0}\mathit{}In}{\mathit{2}r}\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{{B}_{\mathrm{H}}\mathrm{tan}\theta \times 2r}{{\mathrm{\mu}}_{0}\mathrm{I}}\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{3.6\times {10}^{-5}\times 2\times 1\times {10}^{-1}}{4\mathrm{\pi}\times {10}^{-7}\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}\Rightarrow n=0.57332\times {10}^{3}=573$

#### Page No 278:

#### Question 17:

Given:

Horizontal component of Earth's magnetic field, *B*_{H} = 3.6 × 10^{−5} T

Deflection shown by the tangent galvanometer,* θ *= 45°

Current through the galvanometer, *I* = 10 mA = 10^{−2} A

Radius of the coil, *r *= 10 cm = 0.1 m

Number of turns in the coil, *n* = ?

We know,

${B}_{\mathrm{H}}\mathrm{tan}\theta =\frac{{\mathrm{\mu}}_{0}\mathit{}In}{\mathit{2}r}\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{{B}_{\mathrm{H}}\mathrm{tan}\theta \times 2r}{{\mathrm{\mu}}_{0}\mathrm{I}}\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{3.6\times {10}^{-5}\times 2\times 1\times {10}^{-1}}{4\mathrm{\pi}\times {10}^{-7}\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}\Rightarrow n=0.57332\times {10}^{3}=573$

#### Answer:

Given:

Number of turns in the coil,* n* = 50

Area of the cross section of the coil, *A* = 2 cm × 2 cm = 2 × 2 × 10^{−4} m^{2}

Current flowing through the coil,* I* = 20 × 10^{−3} A

Magnetic field strength due to the presence of the poles, *B* = 0.5 T

The torque experienced by the coil placed in an external magnetic field $\left(\tau \right)$ is given by

$\tau \mathit{=}nI\left(\overrightarrow{A}\mathit{\times}\overrightarrow{B}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \tau \mathit{=}nIAB\mathrm{sin}90\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \tau =50\times 20\times {10}^{-3}\times 4\times {10}^{-4}\times 0.5\phantom{\rule{0ex}{0ex}}\Rightarrow \tau =2\times {10}^{-4}\mathrm{N}-\mathrm{m}$

#### Page No 278:

#### Question 18:

Given:

Number of turns in the coil,* n* = 50

Area of the cross section of the coil, *A* = 2 cm × 2 cm = 2 × 2 × 10^{−4} m^{2}

Current flowing through the coil,* I* = 20 × 10^{−3} A

Magnetic field strength due to the presence of the poles, *B* = 0.5 T

The torque experienced by the coil placed in an external magnetic field $\left(\tau \right)$ is given by

$\tau \mathit{=}nI\left(\overrightarrow{A}\mathit{\times}\overrightarrow{B}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \tau \mathit{=}nIAB\mathrm{sin}90\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \tau =50\times 20\times {10}^{-3}\times 4\times {10}^{-4}\times 0.5\phantom{\rule{0ex}{0ex}}\Rightarrow \tau =2\times {10}^{-4}\mathrm{N}-\mathrm{m}$

#### Answer:

Given:

Deflection in the magnetometer in the given position when placed in the magnetic field of a short magnet, *θ* = 37°

Separation between the magnet and the needle, *d* = 10 cm = 0.1 m

Let *M* be the magnetic moment of the magnet and *B*_{H} be Earth's horizontal magnetic field.

According to the magnetometer theory,

$\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}\frac{{\left({d}^{2}-{l}^{2}\right)}^{2}}{2d}\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{the}\mathrm{short}\mathrm{magnet},\phantom{\rule{0ex}{0ex}}\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}\times \frac{{d}^{4}}{2d}\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow \frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{4\mathrm{\pi}\times {10}^{-7}}\times \frac{{\left(0.1\right)}^{3}}{2}\times \mathrm{tan}37\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{M}{{B}_{H}}=0.5\times 0.75\times 1\times {10}^{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{M}{{B}_{H}}=3.75\times {10}^{3}\mathrm{A}-{\mathrm{m}}^{2}/\mathrm{T}$

#### Page No 278:

#### Question 19:

Given:

Deflection in the magnetometer in the given position when placed in the magnetic field of a short magnet, *θ* = 37°

Separation between the magnet and the needle, *d* = 10 cm = 0.1 m

Let *M* be the magnetic moment of the magnet and *B*_{H} be Earth's horizontal magnetic field.

According to the magnetometer theory,

$\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}\frac{{\left({d}^{2}-{l}^{2}\right)}^{2}}{2d}\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{the}\mathrm{short}\mathrm{magnet},\phantom{\rule{0ex}{0ex}}\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}\times \frac{{d}^{4}}{2d}\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow \frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{4\mathrm{\pi}\times {10}^{-7}}\times \frac{{\left(0.1\right)}^{3}}{2}\times \mathrm{tan}37\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{M}{{B}_{H}}=0.5\times 0.75\times 1\times {10}^{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{M}{{B}_{H}}=3.75\times {10}^{3}\mathrm{A}-{\mathrm{m}}^{2}/\mathrm{T}$

#### Answer:

Given:

Deflection in the magnetometer in the given position when placed in the magnetic field of a short magnet, *θ* = 37°

Separation between the magnet and the needle, *d* = 10 cm = 0.1 m

Let *M* be the magnetic moment of the magnet and *B*_{H} be Earth's horizontal magnetic field.

According to the magnetometer theory,

$\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}\frac{{\left({d}^{2}-{l}^{2}\right)}^{2}}{2d}\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{the}\mathrm{short}\mathrm{magnet},\phantom{\rule{0ex}{0ex}}\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}\times \frac{{d}^{4}}{2d}\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow \frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{4\mathrm{\pi}\times {10}^{-7}}\times \frac{{\left(0.1\right)}^{3}}{2}\times \mathrm{tan}37\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{M}{{B}_{H}}=0.5\times 0.75\times 1\times {10}^{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{M}{{B}_{H}}=3.75\times {10}^{3}\mathrm{A}-{\mathrm{m}}^{2}/\mathrm{T}$

= 3.75 × 10^{3} A-m^{2}/T

Deflection in the magnetometer *θ* = 37°

From the magnetometer theory in Tan-B position, we have

$\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}{\left({d}^{2}+{l}^{2}\right)}^{3/2}\mathrm{tan}\mathit{}\theta $

Since for the short magnet *l* <<* d, *we can neglect *l* w.r.t. *d*.

Now,

$\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}{d}^{3}\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow 3.75\times {10}^{3}=\frac{1}{{10}^{-7}}\times {d}^{\mathit{3}}\times 0.75\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{3}=\frac{3.75\times {10}^{3}\times {10}^{-7}}{0.75}=5\times {10}^{-4}\phantom{\rule{0ex}{0ex}}\Rightarrow d=\sqrt[3]{5\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}\Rightarrow d=0.079\mathrm{m}=7.9\mathrm{cm}$

Magnet will be at 7.9 cm from the centre.

#### Page No 278:

#### Question 20:

Given:

Deflection in the magnetometer in the given position when placed in the magnetic field of a short magnet, *θ* = 37°

Separation between the magnet and the needle, *d* = 10 cm = 0.1 m

Let *M* be the magnetic moment of the magnet and *B*_{H} be Earth's horizontal magnetic field.

According to the magnetometer theory,

$\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}\frac{{\left({d}^{2}-{l}^{2}\right)}^{2}}{2d}\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{the}\mathrm{short}\mathrm{magnet},\phantom{\rule{0ex}{0ex}}\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}\times \frac{{d}^{4}}{2d}\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow \frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{4\mathrm{\pi}\times {10}^{-7}}\times \frac{{\left(0.1\right)}^{3}}{2}\times \mathrm{tan}37\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{M}{{B}_{H}}=0.5\times 0.75\times 1\times {10}^{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{M}{{B}_{H}}=3.75\times {10}^{3}\mathrm{A}-{\mathrm{m}}^{2}/\mathrm{T}$

= 3.75 × 10^{3} A-m^{2}/T

Deflection in the magnetometer *θ* = 37°

From the magnetometer theory in Tan-B position, we have

$\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}{\left({d}^{2}+{l}^{2}\right)}^{3/2}\mathrm{tan}\mathit{}\theta $

Since for the short magnet *l* <<* d, *we can neglect *l* w.r.t. *d*.

Now,

$\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}{d}^{3}\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow 3.75\times {10}^{3}=\frac{1}{{10}^{-7}}\times {d}^{\mathit{3}}\times 0.75\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{3}=\frac{3.75\times {10}^{3}\times {10}^{-7}}{0.75}=5\times {10}^{-4}\phantom{\rule{0ex}{0ex}}\Rightarrow d=\sqrt[3]{5\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}\Rightarrow d=0.079\mathrm{m}=7.9\mathrm{cm}$

Magnet will be at 7.9 cm from the centre.

#### Answer:

Given:

Ratio, $\frac{M}{{B}_{H}}=40{\mathrm{Am}}^{2}/\mathrm{T}$

Since the magnet is short, *l* can be neglected.

So, using the formula for $\frac{\mathit{M}}{{\mathit{B}}_{\mathit{H}}}$ from the magnetometer theory and substituting all values, we get

$\mathit{}\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}\frac{{d}^{3}}{2}=40\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{3}=40\times {10}^{-7}\times 2\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{3}=8\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow d=2\times {10}^{-2}\mathrm{m}=2\mathrm{cm}$

Thus, the magnet should be placed in such a way that its north pole points towards the south and it is 2 cm away from the needle.

#### Page No 278:

#### Question 21:

Given:

Ratio, $\frac{M}{{B}_{H}}=40{\mathrm{Am}}^{2}/\mathrm{T}$

Since the magnet is short, *l* can be neglected.

So, using the formula for $\frac{\mathit{M}}{{\mathit{B}}_{\mathit{H}}}$ from the magnetometer theory and substituting all values, we get

$\mathit{}\frac{M}{{B}_{H}}=\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{0}}\frac{{d}^{3}}{2}=40\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{3}=40\times {10}^{-7}\times 2\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{3}=8\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow d=2\times {10}^{-2}\mathrm{m}=2\mathrm{cm}$

Thus, the magnet should be placed in such a way that its north pole points towards the south and it is 2 cm away from the needle.

#### Answer:

Given:

Time taken by the bar magnet to complete one oscillation, *T* = $\frac{\mathrm{\pi}}{10}\mathrm{s}$

Moment of inertia of the magnet about the axis of rotation, *I* = 1.2 × 10^{−4} kgm^{2}

Horizontal component of Earth's magnetic field, *B*_{H} = 30 μT

Time period of oscillating magnetometer $\left(T\right)$ is given by

$T=2\mathrm{\pi}\sqrt{\frac{I}{M{B}_{H}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Here,*M* = Magnetic moment of the magnet

On substituting the respective values, we get

$\frac{\mathrm{\pi}}{10}=2\mathrm{\pi}\sqrt{\frac{1.2\times {10}^{-4}}{\mathrm{M}\times 30\times {10}^{-6}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{1}{20}\right)}^{2}=\frac{1.2\times {10}^{-4}}{M\times 30\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}\Rightarrow M=\frac{1.2\times {10}^{-4}\times 400}{30\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}\Rightarrow M=16\times {10}^{2}\mathrm{A}-{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow M=1600\mathrm{A}-{\mathrm{m}}^{2}$

#### Page No 278:

#### Question 22:

Given:

Time taken by the bar magnet to complete one oscillation, *T* = $\frac{\mathrm{\pi}}{10}\mathrm{s}$

Moment of inertia of the magnet about the axis of rotation, *I* = 1.2 × 10^{−4} kgm^{2}

Horizontal component of Earth's magnetic field, *B*_{H} = 30 μT

Time period of oscillating magnetometer $\left(T\right)$ is given by

$T=2\mathrm{\pi}\sqrt{\frac{I}{M{B}_{H}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Here,*M* = Magnetic moment of the magnet

On substituting the respective values, we get

$\frac{\mathrm{\pi}}{10}=2\mathrm{\pi}\sqrt{\frac{1.2\times {10}^{-4}}{\mathrm{M}\times 30\times {10}^{-6}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{1}{20}\right)}^{2}=\frac{1.2\times {10}^{-4}}{M\times 30\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}\Rightarrow M=\frac{1.2\times {10}^{-4}\times 400}{30\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}\Rightarrow M=16\times {10}^{2}\mathrm{A}-{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow M=1600\mathrm{A}-{\mathrm{m}}^{2}$

#### Answer:

Given:

Number of oscillations per second made by the combination of bar magnets with like poles, ${\nu}_{1}$= 10 s^{$-1$}

Number of oscillations per second made by the combination of bar magnets with unlike poles, ${\nu}_{2}$ = 2 s^{$-1$}

The frequency of oscillations in the magnetometer $\left(\nu \right)$ is given by

$\upsilon =\frac{1}{2\mathrm{\pi}}\sqrt{\frac{M{B}_{\mathrm{H}}}{I}}$

When like poles are tied together, the effective magnetic moment is*M = **M*_{1} − *M*_{2}

When unlike poles are tied together, the effective magnetic moment is*M = M*_{1}* + ${M}_{2}$*

As the frequency of oscillations is directly proportional to the magnetic moment,

$\frac{{\upsilon}_{1}}{{\upsilon}_{2}}=\sqrt{\frac{{M}_{1}-M{\hspace{0.17em}}_{2}}{{M}_{1}+{M}_{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{10}{2}\right)}^{2}=\frac{{M}_{1}-{M}_{2}}{{M}_{1}+{M}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{25}{1}=\frac{{M}_{1}-{M}_{2}}{{M}_{1}+{M}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{25+1}{25-1}=\frac{{M}_{1}-{M}_{2}+{M}_{1}+{M}_{2}}{{M}_{1}-{M}_{2}-{M}_{1}-{M}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{26}{24}=\frac{2{M}_{1}}{-2{M}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{M}_{1}}{{M}_{2}}=-\frac{26}{24}=-\frac{13}{12}$

Hence, the ratio of the effective magnetic moment is $-\frac{13}{12}$.

#### Page No 278:

#### Question 23:

Given:

Number of oscillations per second made by the combination of bar magnets with like poles, ${\nu}_{1}$= 10 s^{$-1$}

Number of oscillations per second made by the combination of bar magnets with unlike poles, ${\nu}_{2}$ = 2 s^{$-1$}

The frequency of oscillations in the magnetometer $\left(\nu \right)$ is given by

$\upsilon =\frac{1}{2\mathrm{\pi}}\sqrt{\frac{M{B}_{\mathrm{H}}}{I}}$

When like poles are tied together, the effective magnetic moment is*M = **M*_{1} − *M*_{2}

When unlike poles are tied together, the effective magnetic moment is*M = M*_{1}* + ${M}_{2}$*

As the frequency of oscillations is directly proportional to the magnetic moment,

$\frac{{\upsilon}_{1}}{{\upsilon}_{2}}=\sqrt{\frac{{M}_{1}-M{\hspace{0.17em}}_{2}}{{M}_{1}+{M}_{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{10}{2}\right)}^{2}=\frac{{M}_{1}-{M}_{2}}{{M}_{1}+{M}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{25}{1}=\frac{{M}_{1}-{M}_{2}}{{M}_{1}+{M}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{25+1}{25-1}=\frac{{M}_{1}-{M}_{2}+{M}_{1}+{M}_{2}}{{M}_{1}-{M}_{2}-{M}_{1}-{M}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{26}{24}=\frac{2{M}_{1}}{-2{M}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{M}_{1}}{{M}_{2}}=-\frac{26}{24}=-\frac{13}{12}$

Hence, the ratio of the effective magnetic moment is $-\frac{13}{12}$.

#### Answer:

Here,

Horizontal component of Earth's magnetic field, *B*_{H} = 24 × 10^{−6} T

Time period of oscillation*, **T*_{1} = 0.1 s

Downward current in the vertical wire,* I* = 18 A

Distance of wire from the magnet, *r* = 20 cm = 0.2 m

In the absence of the wire net magnetic field, *B*_{H} = 24 × 10^{−6} T

When a current-carrying wire is placed near the magnet, the effective magnetic field gets changed.

Now the net magnetic field can be obtained by subtracting the magnetic field due to the wire from Earth's magnetic field.*B* = *B*_{H} −* **B*_{wire}

Thus, the magnetic field due to the current-carrying wire $\left({B}_{wire}\right)$ is given by*B* = $\frac{{\mu}_{0}I}{2\pi r}$

The net magnetic field $\left(B\right)$ is given by

$B=24\times {10}^{-6}-\frac{{\mathrm{\mu}}_{0}\mathit{}I}{2\mathrm{\pi}r}\phantom{\rule{0ex}{0ex}}B=24\times {10}^{-6}-\frac{2\times {10}^{-7}\times 18}{0.2}\phantom{\rule{0ex}{0ex}}B=\left(24-10\right)\times {10}^{-6}\phantom{\rule{0ex}{0ex}}B=14\times {10}^{-6}$

Time period of the coil $\left(T\right)$ is given by

$T=2\mathrm{\pi}\sqrt{\frac{I}{M{B}_{\mathrm{H}}}}\phantom{\rule{0ex}{0ex}}$

Let *T*_{1} and *T*_{2} be the time periods of the coil in the absence of the wire and in the presence the wire respectively.

As time period is inversely proportional to magnetic field,

$\frac{{T}_{1}}{{T}_{2}}=\sqrt{\frac{B}{{B}_{\mathrm{H}}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{0.1}{{T}_{2}}=\sqrt{\frac{14\times {10}^{-6}}{24\times {10}^{-6}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{0.1}{{T}_{2}}\right)}^{2}=\frac{14}{24}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{2}^{2}=\frac{0.01\times 14}{24}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{2}=0.076\mathrm{s}$

#### Page No 278:

#### Question 24:

Here,

Horizontal component of Earth's magnetic field, *B*_{H} = 24 × 10^{−6} T

Time period of oscillation*, **T*_{1} = 0.1 s

Downward current in the vertical wire,* I* = 18 A

Distance of wire from the magnet, *r* = 20 cm = 0.2 m

In the absence of the wire net magnetic field, *B*_{H} = 24 × 10^{−6} T

When a current-carrying wire is placed near the magnet, the effective magnetic field gets changed.

Now the net magnetic field can be obtained by subtracting the magnetic field due to the wire from Earth's magnetic field.*B* = *B*_{H} −* **B*_{wire}

Thus, the magnetic field due to the current-carrying wire $\left({B}_{wire}\right)$ is given by*B* = $\frac{{\mu}_{0}I}{2\pi r}$

The net magnetic field $\left(B\right)$ is given by

$B=24\times {10}^{-6}-\frac{{\mathrm{\mu}}_{0}\mathit{}I}{2\mathrm{\pi}r}\phantom{\rule{0ex}{0ex}}B=24\times {10}^{-6}-\frac{2\times {10}^{-7}\times 18}{0.2}\phantom{\rule{0ex}{0ex}}B=\left(24-10\right)\times {10}^{-6}\phantom{\rule{0ex}{0ex}}B=14\times {10}^{-6}$

Time period of the coil $\left(T\right)$ is given by

$T=2\mathrm{\pi}\sqrt{\frac{I}{M{B}_{\mathrm{H}}}}\phantom{\rule{0ex}{0ex}}$

Let *T*_{1} and *T*_{2} be the time periods of the coil in the absence of the wire and in the presence the wire respectively.

As time period is inversely proportional to magnetic field,

$\frac{{T}_{1}}{{T}_{2}}=\sqrt{\frac{B}{{B}_{\mathrm{H}}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{0.1}{{T}_{2}}=\sqrt{\frac{14\times {10}^{-6}}{24\times {10}^{-6}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{0.1}{{T}_{2}}\right)}^{2}=\frac{14}{24}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{2}^{2}=\frac{0.01\times 14}{24}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{2}=0.076\mathrm{s}$

#### Answer:

Given:

Frequency of oscillations of the bar magnet in the oscillation magnetometer, $\nu $ = 40 min^{$-1$}

Time period for which the bar magnet is placed in the oscillation magnetometer, *T*_{1} = $\frac{1}{40}\mathrm{min}$

The time period of oscillations of the bar magnet $\left(T\right)$ is given by

$\mathit{}T=2\mathrm{\pi}\sqrt{\frac{I}{M{B}_{H}}}\phantom{\rule{0ex}{0ex}}$

Here,*I* = Moment of inertia*M* = Magnetic moment of the bar magnet*B*_{H} = Horizontal component of the magnetic field

Now, let *T*_{2} be the time period for which the second demagnetised magnet is placed over the magnet.

As the second magnet is demagnetised, the combination will have the same values of *M* and *B*_{H} as those for the single magnet. However, variation will be there in the value of* I* on placing the second demagnetised magnet.

$\therefore \frac{{T}_{\mathit{1}}}{{T}_{\mathit{2}}}\mathit{=}\frac{{I}_{\mathit{1}}}{{I}_{\mathit{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{40{T}_{2}}=\sqrt{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{1600{T}_{2}^{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{2}^{2}=\frac{1}{800}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{2}=0.03536\mathrm{min}$

For 1 oscillation,

Time taken = 0.03536 min

For 40 oscillations,

Time taken = 0.03536 × 40

= 1.414 min$=\sqrt{2}\mathrm{min}$

#### Page No 278:

#### Question 25:

Given:

Frequency of oscillations of the bar magnet in the oscillation magnetometer, $\nu $ = 40 min^{$-1$}

Time period for which the bar magnet is placed in the oscillation magnetometer, *T*_{1} = $\frac{1}{40}\mathrm{min}$

The time period of oscillations of the bar magnet $\left(T\right)$ is given by

$\mathit{}T=2\mathrm{\pi}\sqrt{\frac{I}{M{B}_{H}}}\phantom{\rule{0ex}{0ex}}$

Here,*I* = Moment of inertia*M* = Magnetic moment of the bar magnet*B*_{H} = Horizontal component of the magnetic field

Now, let *T*_{2} be the time period for which the second demagnetised magnet is placed over the magnet.

As the second magnet is demagnetised, the combination will have the same values of *M* and *B*_{H} as those for the single magnet. However, variation will be there in the value of* I* on placing the second demagnetised magnet.

$\therefore \frac{{T}_{\mathit{1}}}{{T}_{\mathit{2}}}\mathit{=}\frac{{I}_{\mathit{1}}}{{I}_{\mathit{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{40{T}_{2}}=\sqrt{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{1600{T}_{2}^{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{2}^{2}=\frac{1}{800}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{2}=0.03536\mathrm{min}$

For 1 oscillation,

Time taken = 0.03536 min

For 40 oscillations,

Time taken = 0.03536 × 40

= 1.414 min$=\sqrt{2}\mathrm{min}$

#### Answer:

Here,

Frequency of oscillations, $\nu $_{1} = 40 oscillations/min

Earth's horizontal magnetic field, *B*_{H} = 25 μT

Magnetic moment of the second magnet, *M* = 1.6 A-m^{2}

Distance at which another short magnet is placed,* d *= 20 cm = 0.2 m

(a) For the north pole of the short magnet facing the north, frequency $\left({v}_{1}\right)$ is given by

${v}_{1}=\frac{1}{2\mathrm{\pi}}\sqrt{\frac{M{B}_{H}}{I}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Here,*M* = Magnetic moment of the magnet*I* = Moment of inertia*B*_{H} = Horizontal component of the magnetic field

Now, let *B *be the magnetic field due to the short magnet.

When the north pole of the second magnet faces the north pole of the first magnet, the effective magnetic field $\left({B}_{eff}\right)$ is given by*B*_{effective} = *B*_{H}$-$*B*

The new frequency of oscillations $\left({v}_{2}\right)$ on placing the second magnet is given by

${v}_{\mathit{2}}\mathit{=}\frac{1}{2\mathrm{\pi}}\mathit{}\sqrt{\frac{M\left({B}_{H}\mathit{-}B\right)}{\mathit{1}}}\phantom{\rule{0ex}{0ex}}$

The magnetic field produced by the short magnet $\left(B\right)$ is given by

$B=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{m}{{d}^{\mathit{3}}}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{10}^{-7}\times 1.6}{8\times {10}^{-3}}=20\mathrm{\mu T}\phantom{\rule{0ex}{0ex}}$

Since the frequency is proportional to the magnetic field,

$\frac{{v}_{\mathit{1}}}{{v}_{\mathit{2}}}=\sqrt{\frac{{B}_{H}}{{B}_{H}\mathit{-}B}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{40}{{v}_{2}}=\sqrt{\frac{25}{5}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{40}{{v}_{2}}=\sqrt{5}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{2}=\frac{40}{\sqrt{5}}=17.88\phantom{\rule{0ex}{0ex}}=18\mathrm{oscillations}/\mathrm{min}$

(b) For the north pole facing the south,

${v}_{1}=\frac{1}{2\mathrm{\pi}}\sqrt{\frac{M{B}_{H}}{I}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{2}=\frac{1}{2\mathrm{\pi}}\sqrt{\frac{M\left(B\mathit{+}{B}_{H}\right)}{\mathit{1}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{v}_{1}}{{v}_{2}}=\sqrt{\frac{{B}_{H}}{B\mathit{+}{B}_{H}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{40}{{v}_{2}}=\sqrt{\frac{25}{45}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{2}=\frac{40}{\sqrt{25/45}}=54\mathrm{oscillations}/\mathrm{min}$

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