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 $100 V.$

7   A circular coil of radius $8.0 cm$ and $ 20 $ turns is rotated about its vertical diameter with an angular speed of $50$ rad s $^{-1}$ in a uniform horizontal magnetic field of magnitude $3.0 *10^{-2} T .$ Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance $10\Omega $, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

Solution :

Max induced emf $= 0.603 V$$\\$ Average induced emf $= 0 V$$\\$ Max current in the coil $= 0.0603 A$$\\$ Average power loss $= 0.018 W$ (Power comes from the external rotor)$\\$ Radius of the circular coil, $r = 8 cm = 0.08 m$$\\$ Area of the coil, $A =\pi r^ 2 =\pi*( 0.08 )^2 m ^2$$\\$ Number of turns on the coil, $N = 20$$\\$ Angular speed, $\omega = 50 \ rad / s$$\\$ Magnetic field strength, $B =3 * 10 ^{- 2} T$$\\$ Resistance of the loop, $R = 10 \Omega $$\\$ Maximum induced emf is given as:$\\$

$e=N \omega AB\\ =20*50*\pi * (0.08)^2*3*10^{-2}\\ =0.60 V$$\\$ The maximum emf induced in the coil is $0.603 V.$$\\$ Over a full cycle, the average emf induced in the coil is zero.$\\$ Maximum current is given as:$\\$

$I=\dfrac{e}{R}\\ =\dfrac{0.603}{10}=0.0603A$$\\$ Average power loss due to joule heating:$\\$ $P=\dfrac{eI}{2}\\ =\dfrac{0.603*0.0603}{2}=0.018W$

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.

8   A horizontal straight wire $10 m$ long extending from east to west is falling with a speed of $5.0 m s^{-1}$ , at right angles to the horizontal component of the earth’s magnetic field, $0.30 * 10^{-4} Wb m^{-2}$ .$\\$ (a) What is the instantaneous value of the emf induced in the wire?$\\$ (b) What is the direction of the emf?$\\$ (c) Which end of the wire is at the higher electrical potential?

Solution :

Length of the wire, $l = 10 m$$\\$ Falling speed of the wire, $v = 5.0 m/s$$\\$ Magnetic field strength,$ B = 0.3 * 10 ^{-4} Wb m ^{- 2}$$\\$ (a) Emf induced in the wire,$\\$ $=0.3 * 10 ^{-4 }* 5 * 10$$\\$ $=1.5 * 10^{-3} V$$\\$ (b) Using Fleming’s rule, it can be inferred that the direction of the induced emf is from West to East.$\\$ (c) The eastern end of the wire is at a higher potential.$\\$

9   Current in a circuit falls from $5.0 A$ to $0.0 A$ in $0.1 s$. If an average emf of $200 V$ induced, give an estimate of the self-inductance of the circuit.

Solution :

Initial current, $I_ 1 = 5.0 A$$\\$ Final current,$ I_ 2 = 0.0 A$$\\$ Change in current, $dI = I_ 1 - I_ 2 = 5 A$$\\$ Time taken for the change,$ t = 0.1 s$$\\$ Average emf,$ e = 200 V$$\\$ For self-inductance (L) of the coil, we have the relation for average emf as: $e=L\dfrac{di}{dt}\\ L=\dfrac{e}{(\dfrac{di}{dt})}\\ =\dfrac{200}{\dfrac{5}{0.1}}=4H$$\\$ Hence, the self induction of the coil is $4 H.$

10   A pair of adjacent coils has a mutual inductance of $1.5 H.$ If the current in one coil changes from $0$ to $20 A$ in $0.5 s$, what is the change of flux linkage with the other coil?

Solution :

Mutual inductance of a pair of coils, $\mu = 1.5 H$$\\$ Initial current,$ I _1 = 0 A$$\\$ Final current $I_ 2 = 20 A$$\\$ Change in current, $dI =I _2 -I _1 = 20 - 0 = 20 A$$\\$ Time taken for the change,$ t = 0.5 s$$\\$ Induced emf,$e=\dfrac{d \phi}{dt} \quad ....(1)$$\\$ Where $d\phi $ is the change in the flux linkage with the coil.$\\$ Emf is related with mutual inductance as:$\\$ $e=\mu \dfrac{dI}{dt} \quad ....(2)$$\\$ Equating equations (1) and (2), we get$\\$ $\dfrac{d\phi}{dt}=\mu \dfrac{dI}{dt}\\ d\phi=1.5*(20)\\ =30Wb$$\\$ Hence, the change in the flux linkage is $30 Wb.$

11   A jet plane is travelling towards west at a speed of $1800 km/h$. What is the voltage difference developed between the ends of the wing having a span of $25 m,$ if the Earth’s magnetic field at the location has a magnitude of $5 * 10^{-4} T$ and the dip angle is $30 ^o$ .

Solution :

Speed of the jet plane, $v = 1800 km/h = 500 m/s$$\\$ Wing span of jet plane,$ l = 25 m$$\\$ Earth’s magnetic field strength, $B =5.0 * 10^{- 4} T$$\\$ Angle of dip, $\delta = 30^o$$\\$ Vertical component of Earth’s magnetic field,$\\$ $B _V = B \sin \delta \\ =5 *10^{- 4} \sin 30^o\\ = 2.5*10^{- 4} T$$\\$ Voltage difference between the ends of the wing can be calculated as:$\\$ $e =( B _V )*l * v\\ = 2.5 * 10^{- 4}* 25 *500\\ = 3.125 V$$\\$ Hence, the voltage difference developed between the ends of the wings is $3.125 V.$

12   .

Solution :

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