NCERT Solutions for Class 12 Physics Chapter 5 - Magnetism and Matter

5.1. A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to $4.5\times {10}^{-2}J$ . What is the magnitude of magnetic moment of the magnet?

Answer:

Given,

The angle between axis of bar magnet and external magnetic field, θ = 30°

Magnetic field strength, B = 0.25 T

Torque on the bar magnet, $\tau$ = 4.5 x ${10}^{-2}$ J

We know,

Torque experienced by a bar magnet placed in a uniform magnetic field is:

$\overrightarrow{\tau }=\overrightarrow{m}\times \overrightarrow{B}\Rightarrow |\overrightarrow{\tau }|$= mBsin $\theta$

$m=\frac{\tau }{Bsin\theta }$

$⟹m=\frac{4.5\times {10}^{-2}J}{0.25T\times sin{30}^{\circ }}$

$\therefore$ m = 0.36 $J{T}^{-1}$

Hence, the magnitude of the moment of the Bar magnet is 0.36 $J{T}^{-1}$ .

5.2. A short bar magnet of magnetic moment $m=0.32J{T}^{-1}$ is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Answer:

Given,

Magnetic moment of magnet, m = 0.32 $J{T}^{-1}$

Magnetic field strength, B = 0.15 T

(a) Stable equilibrium: When the magnetic moment is along the magnetic field i.e. $\theta =0^{\circ }$

(b) Unstable equilibrium: When the magnetic moment is at 180° with the magnetic field i.e. $\theta ={180}^{\circ }$

(c) We know that,

U = $-\overrightarrow{m}.\overrightarrow{B}$ = -mBcos $\theta$

By putting the given values:

U = (-0.32)(0.15)(cos $0^{\circ }$ ) = -0.048 J

Therefore, Potential energy of the system in stable equilibrium is -0.048 J

Similarly,

U = (-0.32)(0.15)(cos ${180}^{\circ }$ ) = 0.048 J

Therefore, Potential energy of the system in unstable equilibrium is 0.048J.

5.3 A closely wound solenoid of 800 turns and area of cross-section $2.5\times {10}^{-4}{m}^2$ carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Answer:

In this case the magnetic field is generated along the axis / length of solenoid so it acts as a magnetic bar.

The magnetic moment is calculated as :-

$M=NIA$

or $=800\times 3\times 2.5\times {10}^{-4}$

or $=0.6J{T}^{-1}$

5.4. If the solenoid in Exercise 5.3 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

Answer:

Given,

Magnetic field strength, B = 0.25 T

Magnetic moment, m = 0.6 JT −1

The angle between the axis of the solenoid and the direction of the applied field, $\theta$ = 30°.

We know, the torque acting on the solenoid is:

$\overrightarrow{\tau }=\overrightarrow{m}x\overrightarrow{B}\Rightarrow |\overrightarrow{\tau }|$= mBsin $\theta$
$\Rightarrow |\overrightarrow{\tau }|$= (0.6 $J{T}^{-1}$ )(0.25 T)(sin 30^o)= 0.075 J= 7.5 x ${10}^{-2}$ J

The magnitude of torque is 7.5 x ${10}^{-2}$ J.

5.5 (a) (i) A bar magnet of magnetic moment $1.5J{T}^{-1}$ lies aligned with the direction of a uniform magnetic field of 0.22 T. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction?

Answer:

Given.

Magnetic moment, M= 1.5 $J{T}^{-1}$

Magnetic field strength, B= 0.22 T

Now,

The initial angle between the axis and the magnetic field, ${\theta }_1$ = 0°

Final angle, ${\theta }_2$ = 90°

We know, The work required to make the magnetic moment normal to the direction of the magnetic field is given as:

$W=-MB(cos{\theta }_2-cos{\theta }_1)$

$⟹W=-(1.5)(0.22)(cos{90}^{\circ }-cos{0}^{\circ })=-0.33(0-1)=0.33J$

5.5 (a) (ii) A bar magnet of magnetic moment $1.5TJ{T}^{-1}$ lies aligned with the direction of a uniform magnetic field of 0.22 T. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment opposite to the field direction?

Answer:

The amount of work required for the given condition will be:-

$W=-MB[\cos {\mathrm{\Theta }}_2-\cos {\mathrm{\Theta }}_1]$

$W=-MB[\cos {180}^{\circ }-\cos 0^{\circ }]$

or $W=2MB$

or $W=2\times 1.5\times 0.22$

or $W=0.66J$

5.5 (b) A bar magnet of magnetic moment $1.5J{T}^{-1}$ lies aligned with the direction of a uniform magnetic field of 0.22 T. What is the torque on the magnet in cases (i) and (ii)?

Answer:

For case (i):

$\theta$ = ${\theta }_2$ = 90°

We know, Torque, $\tau =MBsin\theta$
$\tau =(1.5)(0.22)sin{90}^{\circ }=0.33J$

For case (ii):

$\theta$ = ${\theta }_2$ = 180°

We know, Torque,

$\tau =MBsin\theta$
$\tau =(1.5)(0.22)sin{180}^{\circ }=0$

5.6 (a) A closely wound solenoid of 2000 turns and area of cross-section $1.6\times {10}^{-4}{m}^2$ , carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. What is the magnetic moment associated with the solenoid?

Answer:

Given,

Number of turns, N = 2000

Area of the cross-section of the solenoid, $A=1.6$ x ${10}^{-4}{m}^2$

Current in the solenoid, I = 4 A

We know, The magnetic moment along the axis of the solenoid is:

m = NIA

m= (2000)(4 A)(1.6 x ${10}^{-4}{m}^2$ = 1.28 $A{m}^2$

5.7 (a). A short bar magnet has a magnetic moment of $0.48J{T}^{-1}$ . Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on the axis,

Answer:

Given,

The magnetic moment of the bar magnet, m = 0.48 $J{T}^{-1}$

Distance from the centre, d = 10 cm = 0.1 m

We know, The magnetic field at distance d, from the centre of the magnet on the axis is:

$B=\frac{{\mu }_{0}m}{2\pi r^3}$

$\therefore B=\frac{4\pi \times {10}^{-7}\times 0.48}{2\pi (0.1{)}^3}$

$⟹$ $B=$ $0.96\times {10}^{-4}T$

Therefore, the magnetic field on the axis, B = 0.96 G

Note: The magnetic field is along the S−N direction (like a dipole!).

5.7 (b). A short bar magnet has a magnetic moment of $0.48J{T}^{-1}$ Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on the equatorial lines (normal bisector) of the magnet.

Answer:

On the equatorial axis,

Distance,d = 10cm = 0.1 m

We know, the magnetic field due to a bar magnet along the equator is:

$B=-\frac{{\mu }_{0}m}{4\pi d^3}$

$\therefore B=-\frac{4\pi \times {10}^{-7}\times 0.48}{4\pi (0.1{)}^3}$

$⟹$ B = $-0.48\times {10}^{-4}T$

Therefore, the magnetic field on the equatorial axis, B = 0.48 G

The negative sign implies that the magnetic field is along the N−S direction.

7. If the bar magnet in Q.6 is turned around by ${180}^{\circ }$ , where will the new null points be located?

Answer:

Given, d = 14 cm

The magnetic field at a distance d from the centre of the magnet on its axis :

$B={\mu }_{0}m/2\pi d^3$

If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial (perpendicular bisector) line.

The magnetic field at a distance d' from the centre of the magnet on the normal bisector is:

$B={\mu }_{0}m/4\pi d^{\prime 3}$

Equating these two, we get:

$\frac{1}{2d^3}=\frac{1}{4d{}^{\prime 3}}⟹\frac{d{}^{\prime 3}}{d^3}=\frac{1}{2}$

d' = 14 x 0.794 = 11.1cm

The new null points will be at a distance of 11.1 cm on the normal bisector.

8 (a). A short bar magnet of magnetic movement $5.25\times {10}^{-2}J{T}^{-1}$ is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on its normal bisector?

Answer:

Given,

The magnetic moment of the bar magnet, $m=5.25\times$ ${10}^{-2}J{T}^{-1}$

The magnitude of earth’s magnetic field at a place, $H=0.42G=0.42\times$ ${10}^{-4}$ T

The magnetic field at a distance R from the centre of the magnet on the normal bisector is:

$B={\mu }_{0}m/4\pi R^3$

When the resultant field is inclined at 45° with earth’s field, B = H

$B=H=0.42\times$ ${10}^{-4}$ $T$

$R^3={\mu }_{0}m/4\pi B$ = $4\pi \times {10}^{-7}\times 5.25\times {10}^{-2}/4\pi \times 0.42\times {10}^{-4}$

$R^3=12.5\times {10}^{-5}{m}^3$

Therefore, R = 0.05 m = 5 cm

9. (a). Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?

Answer:

At high temperatures, alignment of dipoles gets disturbed due to the random thermal motion of molecules in a paramagnetic sample. But when cooled, this random thermal motion reduces. Hence, a paramagnetic sample displays greater magnetization when cooled.

9. (b). Why is diamagnetism, in contrast, almost independent of temperature?

Answer:

The magnetism in a diamagnetic substance is due to induced dipole moment. So the random thermal motion of the atoms does not affect it which is dependent on temperature. Hence diamagnetism is almost independent of temperature.

9. (c). If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?

Answer:

A toroid using bismuth for its core will have slightly greater magnetic field than a toroid with an empty core because bismuth is a diamagnetic substance.

9. (d). Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?

Answer:

We know that the permeability of ferromagnetic materials is inversely proportional to the applied magnetic field. Therefore it is more for a lower field.

9. (e). Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?

Answer:

Since the permeability of ferromagnetic material is always greater than one, the magnetic field lines are always nearly normal to the surface of ferromagnetic materials at every point.

9. (f). Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?

Answer:

Yes, the maximum possible magnetisation of a paramagnetic sample will be of the same order of magnitude as the magnetisation of a ferromagnet for very strong magnetic fields.

10. (a). Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.

Answer:

According to the graph between B (external magnetic field) and H (magnetic intensity) in ferromagnetic materials, magnetization persists even when the external field is removed. This shows the irreversibility of magnetization in a ferromagnet.

10 (b). The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy?

Answer:

Material that has a greater area of hysteresis loop will dissipate more heat energy. Hence after going through repeated cycles of magnetization, a carbon steel piece dissipates greater heat energy than a soft iron piece, as the carbon steel piece has a greater hysteresis curve area.

10 (c) ‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement.

Answer:

Ferromagnets have a record of memory of the magnetisation cycle. Hence it can be used to store memories.

11 (d). What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer?

Answer:

Ceramic, a ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer.

11 (e). A certain region of space is to be shielded from magnetic fields. Suggest a method.

Answer:

The region can be surrounded by a coil made of soft iron to shield from magnetic fields.

12. A long straight horizontal cable carries a current of 2.5 A in the direction ${10}^{\circ }$ south of west to ${10}^{\circ }$ north of east. The magnetic meridian of the place happens to be ${10}^{\circ }$ west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)

Answer:

Given,

Current in the cable, I = 2.5 A

Earth’s magnetic field at the location, H = 0.33 G = 0.33 × 10^-4 T

The angle of dip, $\delta$ = 0

Let the distance of the line of the neutral point from the horizontal cable = r

The magnetic field at the neutral point due to current carrying cable is:

$H_n={\mu }_{0}I/2\pi r$ ,

We know, Horizontal component of earth’s magnetic field, $H_E$ = $Hcos$ $\delta$

Also, at neutral points, $H_E=H_n$

⇒ $Hcos$ $\delta$ = ${\mu }_{0}I/2\pi r$

$\Rightarrow 0.33\times {10}^{-4}T\cos 0^{\circ }=\frac{4\pi \times {10}^{-7\times 2.5}}{2\pi r}$

$\Rightarrow r=1.515cm$

Required distance is 1.515 cm.

13. A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is ${35}^{\circ }$ . The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

Answer:

Number of long straight horizontal wires = 4

The current carried by each wire = 1A

earth’s magnetic field at the place = 0.39 G

the angle of dip = 35^o

magnetic field due to infinite current-carrying straight wire

$B^{\prime }=\frac{{\mu }_{0}I}{2\pi r}$

r=4cm =0.04 m

$B^{\prime }=\frac{4\pi \times {10}^{-7}\times 1}{2\pi \times 4\times {10}^{-2}}$

Magnetic field due to such 4 wires

$B=4\times \frac{4\pi \times {10}^{-7}\times 1}{2\pi \times 4\times {10}^{-2}}=2\times {10}^{-5}T$

The horizontal component of the earth's magnetic field

$H=0.39\times {10}^{-4}cos35=0.319\times {10}^{-4}T=3.19\times {10}^{-5}T$

The vertical component of the earth's magnetic field

$V=0.39\times {10}^{-4}sin35=0.22\times {10}^{-4}T=2.2\times {10}^{-5}T$

At the point below the cable

$H^{\prime }=H-B=3.19\times {10}^{-5}-2\times {10}^{-5}=1.19\times {10}^{-5}T$

The resulting field is

$\sqrt{H^{\prime 2}+V^2}=\sqrt{(1.19\times {10}^{-5}{)}^2+(2.2\times {10}^{-5}{)}^2}=2.5\times {10}^{-5}T=0.25G$

14 (a). A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45 degree with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east. Determine the horizontal component of the earth’s magnetic field at the location.

Answer:

Given,

Number of turns in the coil, n = 30

Radius of coil, r = 12cm = 0.12m

Current in the coil, I = 0.35A

The angle of dip, $\delta$ = 45^o

We know, Magnetic fields due to current carrying coils, B = ${\mu }_{0}nI/2r$

$B=$ $4\pi \times {10}^{-7}\times 30\times 0.35/2\times 0.12$= 5.49 $\times$ ${10}^{-5}$ $T$

Now, Horizontal component of the earth’s magnetic field B_H= Bsin $\delta$

$B_H= 5.49 \times10^{-5}T sin45^o$

$B_H=3.88\times10^{-5}T$ (Hint: Take sin45^o as 0.7)

$B_H=0.388 G$

14 (b). A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of ${45}^{\circ }$ with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east. The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of ${90}^{\circ }$ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Answer:

When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of ${90}^{\circ }$ in the anticlockwise sense looking from above, then the needle will reverse its direction. The new direction will be from east to west.

15. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is ${60}^{\circ }$ , and one of the fields has a magnitude of $1.2\times {10}^{-2}$ T. If the dipole comes to stable equilibrium at an angle of ${15}^{\circ }$ with this field, what is the magnitude of the other field?

Answer:

Given,

The magnitude of the first magnetic field, B₁ = 1.2 × 10^–2 T

The angle between the magnetic field directions, $\theta$ = 60°

The angle between the dipole and the magnetic field $B_1$ is ${\theta }_1$ = 15°

Let B 2 be the magnitude of the second magnetic field and M be the magnetic dipole moment

Therefore, the angle between the dipole and the magnetic field B₂ is ${\theta }_2$ = $\theta -{\theta }_1$ = 45°

Now, at rotational equilibrium,

The torque due to field B₁ = Torque due to field B₂

$M{B}_{1}sin{\theta }_1=M{B}_{2}sin{\theta }_2$

$B_2=\frac{M{B}_{1}sin{\theta }_1}{Msin{\theta }_2}=\frac{1.2\times {10}^{-2}\times sin{15}^{\circ }}{sin{45}^{\circ }}$

$B_2=4.39\times$ ${10}^{-3}$ $T$

Hence the magnitude of the second magnetic field $=4.39\times$ ${10}^{-3}$ $T$

16. A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm $(m_e=9.11\times {10}^{-31}Kg)$ . [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]

Answer:

The energy of electron beam = 18 eV

We can write:-

$E=\frac{1}{2}m{v}^2$

so $v=\sqrt{\frac{2E}{M}}$

We are given horizontal magentic field : B = 0.40 G

Also, $Bev=\frac{m{v}^2}{r}$

We obtain, $r=\frac{1}{Be}\sqrt{2EM}$

or $r=11.3m$

Using geometry, we can write:-

$\sin \mathrm{\Theta }=\frac{x}{r}=\frac{0.3}{11.3}$

and $y=r-r\cos \mathrm{\Theta }$

or $y=r(1-\sqrt{1-{\sin }^2\mathrm{\Theta }})$

or $y\approx 4mm$

17. A sample of paramagnetic salt contains $2.0\times {10}^{24}$ atomic dipoles each of dipole moment $1.5\times {10}^{-23}J{T}^{-1}$ The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to $15\mathrm{\%}$ . What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)

Answer:

Given,

Magnetic field, $B_1$ = 0.64 T

Temperature, ${\theta }_1$ = 4.2K

And, saturation = 15%

Hence, Effective dipole moment, $M_1$ = 15% of Total dipole moment

$M_1$ = 0.15 x (no. of atomic dipole × individual dipole moment)

$M_1$ = $0.15\times 2\times {10}^{24}\times 1.5\times {10}^{-23}$ = 4.5 $J{T}^{-1}$

Now, Magnetic field, $B_2$ = 0.98 T and Temperature, ${\theta }_2$ = 2.8 K

Let $M_2$ be the new dipole moment.

We know that according to Curie’s Law, $M\propto \frac{B}{\theta }$

∴ The ratio of magnetic dipole moments

$\frac{M_2}{M_1}=\frac{B_2\times {\theta }_1}{B_1\times {\theta }_2}$

$⟹M_2=\frac{B_2\times {\theta }_1}{B_1\times {\theta }_1}\times M_1$

$⟹M_2=\frac{0.98T\times 4.2K}{0.64T\times 2.8K}\times 4.5J{T}^{-1}$ $=10.336$ $J{T}^{-1}$

Therefore, the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K = 10.336 $J{T}^{-1}$

18. A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?

Answer:

Given,

Radius of ring, r = 15cm = 0.15m

Number of turns in the ring, n = 3500

Relative permeability of the ferromagnetic core, ${\mu }_r$ = 800

Current in the Rowland ring, I = 1.2A

We know,

Magnetic Field due to a circular coil, B = $\frac{{\mu }_r{\mu }_{0}nI}{2\pi r}$

∴ B = $\frac{4\pi \times {10}^{-7}\times 800\times 3500\times 1.2}{2\pi \times 0.15}$ = 4.48T

Therefore, the magnetic field B in the core for a magnetising current is 4.48 T

19. The magnetic moment vectors ${\mu }_s$ and ${\mu }_l$ associated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron, are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by: ${\mu }_s=-(e/m)S.{\mu }_l=-(e/2m)l$. Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

Answer:

We know,

${\mu }_l=-(\frac{e}{2m})l$

$\therefore {\mu }_l=-(\frac{e}{2m})l$ is in expected from classical physics.

Now, the magnetic moment associated with the orbital motion of the electron is:

${\mu }_l$ = Current x Area covered by orbit = I x A= $(\frac{e}{T})\pi r^2$

And, l = angular momentum = mvr = $m(\frac{2\pi r}{T})r$

(m is the mass of the electron having charge (-e), r is the radius of the orbit of by the electron around the nucleus and T is the time period.)

Dividing these two equations:

$\frac{{\mu }_l}{l}=-\frac{e}{T}\pi r^2\times \frac{T}{m\times 2\pi r^2}=-\frac{e}{2m}$

${\mu }_l=(-\frac{e}{2m})l$ , which is the same result predicted by quantum theory.

The negative sign implies that ${\mu }_l$ and l are anti-parallel.