Physics Ncert Solutions Class 12th

6.4 Length of the rectangular wire loop, l = 8 cm = 0.08 m

Width of the rectangular loop, b = 2 cm = 0.02 m

Hence, the area of the rectangular loop, A = l $*b$ = 0.08 $*0.02$ = 1.6 $*{10}^{-3}$ $m^2$

Magnetic field strength, B= 0.3 T

Velocity of the loop, v = 1 cm/s = 0.01 m /s

Emf developed in the longer side:

Emf developed in the loop is given by e = Blv = 0.3 $*0.08*0.01$ = 2.4 $*{10}^{-4}$ V

Time taken to travel the width = $\frac{Distancetravelled}{velocity}$ = $\frac{b}{v}$ = $\frac{0.02}{0.01}$ s= 2 s

Hence the induced voltage is 2.4 $*{10}^{-4}$ V, which lasts for 2 s.

Emf developed in the shorter side:

Emf developed in the loop is given by e = Bbv = 0.3 $*0.02*0.01$ = 0.6 $*{10}^{-4}$ V

Time taken to travel the width = $\frac{Distancetravelled}{velocity}$ = $\frac{l}{v}$ = $\frac{0.08}{0.01}$ s= 8 s

Hence the induced voltage is 0.6 $*{10}^{-4}$ V, which lasts for 8 s.

6.3 Number of turns on the solenoid = 15 turns per cm = 1500 turns per m

Hence, number of turns per unit length, n = 1500

The loop area in the solenoid, A = 2.0 ${cm}^2$ = 2 $*{10}^{-4}$ $m^2$

Current carrying by the solenoid, changes from 2 .0 A to 4.0 A

So, change of current, di = 4 – 2 = 2 A

Change in time, dt = 0.1 s

According to Faraday's law, the induced emf, e = $\frac{d\phi }{dt}$ …………. (i)

Where $\phi$ = induced flux through the small loop = BA …………. (ii)

Magnetic field is given by B = ${\mu }_{0}ni$ ……………………………. (iii)

Where ${\mu }_0$ = Permeability of free space = 4 $\pi *{10}^{-7}$ H/m

From equation (i),

e = $\frac{d}{dt}\left(BA\right)$

= A ${\mu }_{0}n*\frac{di}{dt}$

= 2 $*{10}^{-4}*$ 4 $\pi *{10}^{-7}*1500*\frac{2}{0.1}$

= 7.54 $*{10}^{-6}$ V

Hence the induced emf while current is changing is 7.54 $*{10}^{-6}$ V

6.2 (a) As the loop changes from irregular to circular shape, its area increases. Hence the magnetic flux linked with it also increases. According to Lenz's law, the induced current should produce magnetic flux in the opposite direction of the original flux. For this induced current should flow in the anti-clockwise direction, i.e. adcb.

(b) As this circular loop is being deformed into a narrow straight wire, its area decreases. The magnetic flux linked also decreases. By Lenz's law, the induced current should produce a flux in the direction of the original flux. For this the induced current should flow in the anti-clock wise direction, i.e. a'd'c'b'.

6.1

The direction of the induced current in a closed loop is given by Lenz's law. The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.

Using Lenz's rule, the direction of the induced current in the given situation can be predicted as follows:

The direction of the induced current is along 'qrpq'.

The direction of the induced current is along 'prqp'.

The direction of the induced current is along 'yzxy'.

The direction of the induced current is along 'zyxz'.

The direction of the induced current is along 'xryx'.

No current is induced since the field lines are lying in the plane of the closed loop.

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.

5.22 Energy of an electron beam, E = 18 keV = 18 $*{10}^3$ eV

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

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

Magnetic field, B = 0.04 G

Mass of an electron, $m_e$ = 9.11 $*{10}^{-31}$ 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 = $\sqrt{\frac{2E_T}{m}}$ = $\sqrt{\frac{2*18\mathrm{}*{10}^3*1.6*\mathrm{}{10}^{-19}}{9.11\mathrm{}*{10}^{-31}}}$ = 79.51 $*{10}^6$ 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 = $\frac{mv}{Be}$ = $\frac{9.11\mathrm{}*{10}^{-31}*79.51\mathrm{}*{10}^6}{0.4*{10}^{-4}*1.6*\mathrm{}{10}^{-19}}$ = 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.