# Electromagnetic Induction

## Physics

### NCERT

1   Predict the direction of induced current in the situations described by the following figures (a) to (f)

##### Solution :

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 situations can be predicted as follows:$\\$ a) The direction of the induced current is along qrpq.$\\$ b) The direction of the induced current is along prqp.$\\$ c) The direction of the induced current is along yzxy.$\\$ d) The direction of the induced current is along zyxz.$\\$ e) The direction of the induced current is along xryx.$\\$ f) No current is induced since the field lines are lying in the plane of the closed loop.

2   Predict the direction of induced current in the situations described by the following figures (a) to (f).

##### Solution :

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3   Use Lenz’s law to determine the direction of induced current in the situations described by Figure:$\\$ (a) A wire of irregular shape turning into a circular shape;$\\$ (b) A circular loop being deformed into a narrow straight wire.$\\$

##### Solution :

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According to Lenz’s law, the direction of the induced emf is such that it tends to produce a current that opposes the change in the magnetic flux that produced it.$\\$ (a) The wire is expanding to form a circle, which means that force is acting outwards on each part of wire because of magnetic field (acting in the downwards direction). The direction of induced current should be such that it will produce magnetic field in upward direction (towards the reader). Hence force on wire will be towards inward direction, i.e. induced current is flowing in anticlockwise direction in the loop from cbad.$\\$ (b) When the shape of a circular loop is deformed into a narrow straight wire, the flux piercing the surface decreases. Hence, the induced current flows along abcd.

4   A long solenoid with $15$ turns per cm has a small loop of area $2.0 cm^2$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from $2.0 A$ to $4.0 A$ in $0.1 s$, what is the induced emf in the loop while the current is changing?

##### Solution :

Number of turns on the solenoid $= 15$ turns/cm $= 1500$ turns/m $\\$ Number of turns per unit length, $n = 1500$ turns$\\$ The solenoid has a small loop of area, $A =2.0 cm^2 = 2 * 10 ^{- 4} m ^2$$\\ Current carried by the solenoid changes from 2 A to 4 A. \therefore Change in current in the solenoid, di = 4 - 2 = 2 A Change in time, dt = 0.1 s$$\\$ Induced emf in the solenoid is given by Faraday’s law as:$\\$ $e=\dfrac{d \phi}{dt}...(i)$$\\ Where,\\ \phi= Induced flux through the small loop = BA ... (ii)\\ Hence, equation (i) reduces to:\\ e=\dfrac{d}{dt}(BA)\\ =A\mu_0n*(\dfrac{di}{dt})\\ =2*10^{-4} * 4 \pi * 10^{-7} * 1500 * \dfrac{2}{0.1}\\ =7.54* 10^{-6}V$$\\$ Hence, the induced voltage in the loop is $=7.54*10^{-6} V$

5   A rectangular wire loop of sides $8 cm$ and $2 cm$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3 T$ directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is $1 cm s^{-1}$ in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

##### Solution :

Length of the rectangular wire, $l = 8 cm = 0.08 m$$\\ Width of the rectangular wire, b = 2 cm = 0.02 m$$\\$ Hence, area of the rectangular loop, $A = lb\\ = 0.08 * 0.02\\ = 1 6 * 10^{- 4} m^2$$\\ Magnetic field strength, B = 0.3 T Velocity of the loop, v = 1 cm/s = 0.01 m/s$$\\$ (a) Emf developed in the loop is given as:$\\$ $e = Blv\\ = 0.3*0.08* 0.01 = 2.4 * 10^{-4 }V$$\\ Time taken to travel along the width,\\ t=\dfrac{\text{Distance travelled}}{\text{Velocity}}=\dfrac{b}{v}\\ =\dfrac{0.02}{0.01}=2s$$\\$ Hence, the induced voltage is $2.4 * 10^{- 4} V$ which lasts for $2 s.$$\\ (b) Emf developed, e = Bbv$$\\$ $= 0.3 *0.02 * 0.01 = 0.6 * 10^{- 4} V$$\\ Time taken to travel along the length,\\ t=\dfrac{\text{Distance travelled}}{\text{Velocity}}=\dfrac{l}{v}\\ =\dfrac{0.08}{0.01}=8s$$\\$ Hence, the induced voltage is $0.6 × 10^{-4} V$ which lasts for $8 s.$

6   A $1.0 m$ long metallic rod is rotated with an angular frequency of $400 \text{rad} s^{-1}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of $0.5 T$ parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

##### Solution :

Length of the rod,$l = 1 m$$\\ Angular frequency, \omega= 400 \text{rad} / s$$\\$ Magnetic field strength, $B = 0.5 T$$\\ One end of the rod has zero linear velocity, while the other end has a linear velocity of l \omega .$$\\$ Average linear velocity of the rod,$v=(\dfrac{l\omega +0}{2})\\ =\dfrac{l\omega}{2}$$\\ Emf developed between the centre and the ring,\\ e=Blv=Bl(\dfrac{l \omega}{2}) =(\dfrac{Bl^2 \omega}{2})\\ =\dfrac{0.5*(1)^2*400}{2}=100V$$\\$ Hence, the emf developed between the centre and the ring is $100 V.$