Class 12th

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.

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 $*{10}^{-4}$ 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 $*\frac{{\mu }_{0}I}{2\pi r}$

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

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

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

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

$H_V$ = H $\sin ?\beta$ = 0.39 $*{10}^{-4}$ = 0.22 $*{10}^{-4}$ 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 $*\frac{{\mu }_{0}I}{2\pi r}$

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

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

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

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

$H_V$ = H $\sin ?\beta$ = 0.39 $*{10}^{-4}\sin ?35°$ = 0.22 $*{10}^{-4}$ 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 $\frac{{\mu }_{0}2\pi nI}{4\pi r}$