Electromagnetic Induction

6.6 Given,

Radius of the circular coil, r = 8.0 cm = 0.08 m

Area of the coil, A = $\pi r^2$ = $\pi *{0.08}^2$ = 20.106 $*{10}^{-3}$ $m^2$

Number of turns in the coil, n = 20

Angular speed, $\omega$ = 50 rad/s

Magnetic field strength, B = 3.0 $*{10}^{-2}$ T

Resistance of the loop, R = 10 Ω

Maximum induced emf is given as e = n $\omega AB$ = 20 $*50*$ 20.106 $*{10}^{-3}*$ 3.0 $*{10}^{-2}$ = 0.603 V

Over a full cycle, the average emf induced in the coil is zero.

Maximum current is given by, $I_{max}$ = $\frac{e}{R}$ = $\frac{0.603}{10}$ = 0.0603 A

Average power loss due to Joule heating is given by, $P_{loss}$ = $\frac{e{I}_{max}}{2}$ = 0.018 W

The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent; it must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.

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.