NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter: Topics Covered, Question with Solutions PDF

Therefore, a set of neutral points parallel to and above the cable are located at a normal distance of 1.52 cm.

 

Q.5.19 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°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

Ans.5.19 Number of horizontal wires in the telephone cable, n = 4

Current in each wire, I = 1.0 A

The Earth’s magnetic field at a location, H = 0.39 G = 0.39 $\times {10}^{-4}$ T

Angle of dip at the location, $\beta$ = 35 $°$

Angle of declination, $\theta \approx$ 0 $°$

For a point 4.0 cm below the cable, r = 4.0 cm = 0.04 m

The horizontal component of earth’s magnetic field can be written as

$H_H$ = H $\cos \beta$ - B, where

B = Magnetic field at 4 cm due to current I in four wires = 4 $\times \frac{{\mu }_{0}I}{2\pi r}$

${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

Then B = 4 $\times \frac{4\pi \times {10}^{-7}\times 1}{2\pi \times 0.04}$ = 2 $\times {10}^{-5}$ T = 0.2 $\times {10}^{-4}$ T = 0.2 G

$H_H$ = H $\cos \beta$ – B = 0.39 $\times {10}^{-4}$ $\cos \beta$ - 0.2 $\times {10}^{-4}$ = 0.12 $\times {10}^{-4}$ T= 0.12 G

The vertical component of Earth’s magnetic field is given as

$H_V$ = H $\sin \beta$ = 0.39 $\times {10}^{-4}$ = 0.22 $\times {10}^{-4}$ T = 0.22 G

The angle made by the field with its horizontal component is given as:

$\theta ={\tan }^{-1}\frac{H_V}{H_H}$ = ${\tan }^{-1}\frac{0.22}{0.12}$ = 61.39 $°$

The resultant magnetic field at the point is given as

H = $\sqrt{{H_H}^2+{H_V}^2}$ = $\sqrt{{0.12}^2+{0.22}^2}$ = 0.25 G

For a point 4.0 cm above the cable, r = 4.0 cm = 0.04 m

The horizontal component of earth’s magnetic field can be written as

$H_H$ = H $\cos \beta$ + B, where

B = Magnetic field at 4 cm due to current I in four wires = 4 $\times \frac{{\mu }_{0}I}{2\pi r}$

${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

Then B = 4 $\times \frac{4\pi \times {10}^{-7}\times 1}{2\pi \times 0.04}$ = 2 $\times {10}^{-5}$ T = 0.2 $\times {10}^{-4}$ T = 0.2 G

$H_H$ = H $\cos \beta$ + B = 0.39 $\times {10}^{-4}$ + 0.2 $\times {10}^{-4}$ = 0.52 $\times {10}^{-4}$ T= 0.52 G

The vertical component of Earth’s magnetic field is given as

$H_V$ = H $\sin \beta$ = 0.39 $\times {10}^{-4}\sin 35°$ = 0.22 $\times {10}^{-4}$ T = 0.22 G

The angle made by the field with its horizontal component is given as:

$\theta ={\tan }^{-1}\frac{H_V}{H_H}$ = ${\tan }^{-1}\frac{0.22}{0.52}$ = 22.93 $°$

The resultant magnetic field at the point is given as

H = $\sqrt{{H_H}^2+{H_V}^2}$ = $\sqrt{{0.52}^2+{0.22}^2}$ = 0.56 G $\frac{{\mu }_{0}2\pi nI}{4\pi r}$

 

Q.5.20 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° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.

(a) Determine the horizontal component of the earth’s magnetic field at the location.

(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Ans.5.20 Number of turns in the circular coil, n = 30

Radius of the circular coil, r = 12 cm = 0.12 m

Current in the coil, I = 0.35 A

Angle of dip, $\beta$ = 45 $°$

The magnetic field due to current I at a distance r is given by

B = $\frac{{\mu }_{0}2\pi nI}{4\pi r}$ , where $B_1$

${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

Then B = $\frac{4\pi \times {10}^{-7}\times 2\pi \times 30\times 0.35}{4\pi \times 0.12}=5.50\times {10}^{-5}$ T

Angle between two fields, $\theta$ = 60 $°$

At stable equilibrium, the angle between the dipole and the field , $B_1$ = ${\theta }_1$ = $15°$

Angle between the dipole and the field $B_2$ , ${\theta }_2$ = $\theta$ - ${\theta }_1$ = 60 $°$ - $15°$ = $45°$

At rotational equilibrium, the torques between both the fields must balance each other.

i.e. Torque due to field $B_1$ = Torque due to field $B_2$

M $B_1\sin {\theta }_1$ = M $B_2\sin {\theta }_2$

$B_2=\frac{B_1\sin {\theta }_1}{\sin {\theta }_2}$ = $\frac{1.2\mathrm{}\times {10}^{-2}sin15°}{\sin 45°}$ =4.39 $\times {10}^{-3}$ T

 

Q.5.22 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

(me = 9.11 × 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.]

Ans.5.22 Energy of an electron beam, E = 18 keV = 18 $\times {10}^3$ eV

Charge of an electron, e = 1.6 $\times$ ${10}^{-19}$ C

Total energy of the electron beam, $E_T$ = E $\times e$ = 18 $\times {10}^{-3}$ $\times$ 1.6 $\times$ ${10}^{-19}$

Magnetic field, B = 0.04 G

Mass of an electron, $m_e$ = 9.11 $\times {10}^{-31}$ kg

Distance up to which the electron beam travels, d = 30 cm = 0.3 m

The kinetic energy of an electron beam, $E_k$ = $\frac{1}{2}$ m $v^2$ = $E_T$

v = $\sqrt{\frac{2E_T}{m}}$ = $\sqrt{\frac{2\times 18\mathrm{}\times {10}^3\times 1.6\times \mathrm{}{10}^{-19}}{9.11\mathrm{}\times {10}^{-31}}}$ = 79.51 $\times {10}^6$ m/s

The electron beam deflects along a circular path of radius r.

The force due to the magnetic field balances the centripetal force of the path

Bev = $\frac{m{v}^2}{r}$ or r = $\frac{mv}{Be}$ = $\frac{9.11\mathrm{}\times {10}^{-31}\times 79.51\mathrm{}\times {10}^6}{0.4\times {10}^{-4}\times 1.6\times \mathrm{}{10}^{-19}}$ = 11.3 m

Let the up and down deflection of the electron beam be x = r (1- cos) where $\theta$ = Angle of declination

$\sin \theta$ = $\frac{d}{r}$ = $\frac{0.3}{11.3}$

$\theta =1.521°$

x = 11.3 (1 – cos 1.521 $°$ ) = 3.98 mm

Therefore up and down deflection of the beam is 3.98 mm.

 

Q.5.23 A sample of paramagnetic salt contains 2.0 × 1024 atomic dipoles each of dipole moment 1.5 × 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%. 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)

Ans.5.23 Number of atomic dipoles, n = 2 $\times {10}^{24}$

Dipole moment of each atomic dipole, M = 1.5 $\times {10}^{-23}$ J/T

Magnetic field, $B_1$ = 0.64 T

The sample is cooled to the temperature, $T_1$ = 4.2 K

Total dipole moment of the atomic dipole, $M_{tot}$ = n $\times$ M

= 2 $\times {10}^{24}$ $\times 1.5\mathrm{}\times {10}^{-23}\mathrm{}=\mathrm{}\mathrm{}30\mathrm{}\mathrm{}\mathrm{J}/\mathrm{T}$

Magnetic saturation is achieved at 15 %

Hence, effective dipole moment, = 15 % $\times 30$ = 4.5 J/T

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

Therefore, the magnetic field in the core is 4.48 T

 

Q.5.25 The magnetic moment vectors and 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:

${\mathit{\mu }}_{\mathit{s}}=-\left(\frac{\mathit{e}}{\mathit{m}}\right)\mathit{s}\dots \dots ..\left(1\right)$

${\mathit{\mu }}_{\mathit{l}}=-\left(\frac{\mathit{e}}{2\mathit{m}}\right)\mathit{l}$ ……..(2)

Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

Ans.5.25 Of the two, the relation, the relation (2) is in accordance with classical physics. It follows easily from the definitions of

${\mathit{\mu }}_{\mathit{l}\mathit{}}$ = IA = $\frac{e}{T})\pi r^2$ ( …..(i)

I = mvr = m(2 $\pi r^2/T)$ ……(ii)

where r is the radius of the circular orbit which the electron of mass m and charge (-e) completes in time T.

Dividing (i) by (ii),

Clearly, $\frac{{\mu }_l}{I}$ = [(e/T) $\pi r^2]$ /[m(2 $\pi r^2/T)$ = -(e/2m)

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

Since the charge of the electron is negative, it is easily seen that ${\mu }_l$ and l are antiparallel, both normal to the plane of the orbit.

Note ${\mathit{\mu }}_{\mathit{s}}\mathit{}$ /s in contrast to ${\mu }_l$ /l is e/m, i.e. twice the classically expected value. This latter result (verified experimentally) is an outstanding consequence of modern quantum theory and cannot be obtained classically.