Chapter 6: Electromagnetic Induction

Q6.21Written answer

There are two coils A and B separated by some distance. If a current of 2 A flows through A, a magnetic flux of 10-2 Wb passes through B (no current through B). If no current passes through A and a current of 1 A passes through B, what is the flux through A?

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Worked solution

Ia is the current passing through the coil

Mab is the mutual induction between A and B

Na is the number of turns in coil A

Nb is the number of turns in coil B

ϕa is the flux linked to coil A due to coil B

ϕb is the flux linked to coil B due to coil A

Total flux is B = M2ϕ2 = MabI1=5mH

Total flux through A = Maϕa = MabI1 = 5mWb

Long Answers

Q6.22Written answer

A magnetic field B = Bo sin ωt k covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d. The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity v, what is the current in the circuit? What is the force needed to keep the wire moving at constant velocity?

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Worked solution

Let wire AB at t = 0 move with velocity v.

At t, x(t) = vt

Motional emf across AB = e1 = Blv

e1 = (B_o sin ꞷt)vd(-j)

e2 = d(ϕ_B)/dt

e2 = -B0 ꞷ cos ꞷtx (t)d

Total emf in the circuit = emf due to change in field + motional emf across AB

Electric current in the clockwise direction is given as = Bod/R (ꞷx cos ꞷt + v sin ꞷt)

Therefore, external force is given as, F_ext = Bo²d²/R (ꞷx cos ꞷt + v sin ꞷt)(sin ꞷt (i)

Q6.23Written answer

A conducting wire XY of mass m and negligible resistance slides smoothly on two parallel conducting wires. The closed-circuit has a resistance R due to AC. AB and CD are perfect conductors. There is a magnetic field B = B t(k ).

(i) Write down the equation for the acceleration of the wire XY.

(ii) If B is independent of time, obtain v(t) , assuming v (0) = u0.

(iii) For (b), show that the decrease in kinetic energy of XY equals the heat lost in R.

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Worked solution

The magnetic flux linked with the loop is given as ϕ_m = B.A = BA cos θ

Where A is the area vector and B is the magnetic field vector.

Emf induced due to change in magnetic field = e1 = -dB(t)/dt lx (t)

Emf induced due to motion = e2 = B(t) lv (t) (-j)

Total emf in the circuit = emf due to change in the field + the motional emf across XY

E = -dB(t)/dt lx (t) – B(t)lv(t)

We know current, I = E/R

Force = IB(t)/R[-dB(t)/dt Ix(t) – I²B²(t)/R dx/dt]

B is independent of the time

Power consumption is given as P = I²R

Therefore, energy consumed in time interval dt is given as = m/2 u0² – m/2 v²(t)

The above equation shows that there is a decrease in the kinetic energy.

Q6.24Written answer

ODBAC is a fixed rectangular conductor of negligible resistance (CO is not connected) and OP is a conductor which rotates clockwise with an angular velocity ω. The entire system is in a uniform magnetic field B whose direction is along the normal to the surface of the rectangular conductor ABDC. The conductor OP is in electric contact with ABDC. The rotating conductor has a resistance of λ per unit length. Find the current in the rotating conductor, as it rotates by 180°.

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Worked solution

i) When the position of the rotating conductor is assumed to be at the time interval t = 0 to t = π/4ꞷ

We get current I = 1/2 Bl²ꞷ/λl sec² ꞷt cos ꞷt = Blꞷ/2λ cos ꞷt

ii) When the position of the rotating conductor is at time interval π/4ꞷ < t < 3π/4ꞷ

We get current I = 1/2 Blꞷ/λ sin ꞷt

iii) When the position of the rotating conductor is at time interval 3π/4ꞷ < t < π/ꞷ

We get current I = 1/2 Blꞷ/ λ sin ꞷt

Q6.25Written answer

Consider an infinitely long wire carrying a current I (t ), with dI dt =λ= constant. Find the current produced in the rectangular loop of wire ABCD if its resistance is R.

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Worked solution

The width of the strip is dr and the length is l which is inside the rectangular box at a distance r from the surface of the current-carrying conductor. The magnetic field across the strip length is given as

l = B(r) = μoI/2πr

B(r) is perpendicular to the paper upward.

Flux in the strip = ϕ = μoIl/2π [log_er]

ε=-dϕ/dt

Therefore, I = μ_oλl/2πR log_ex/x_o

Q6.26Written answer

A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current II ( ) t = o (1– /t T ) for 0 ≤ ≤ t T and I (0) = 0 for t > T. Find the total charge passing through a given point in the loop, in time T. The resistance of the loop is R.

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Worked solution

If t is the instantaneous current then,

I(t) = 1/R dϕ/dt

If q is the charge passing in time t

I(t) = dQ/dt

dQ/dt = 1/R dϕ/dt

Integrating the equation we get,

Q = μ₀L1L2/2πR log (L2 + x/x)

Q6.27Written answer

A magnetic field B is confined to a region r a ≤ and points out of the paper (the z-axis), r = 0 being the centre of the circular region. A charged ring (charge = Q) of radius b, b > a and mass m lies in the x-y plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time ∆t. Find the angular velocity ω of the ring after the field vanishes.

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Worked solution

The magnetic flux across the conducting ring reduces to zero from maximum when the magnetic field is reduced in ∆t.

Induced emf = E2πb

From Faraday’s law of emf,

The induced emf = rate of change of magnetic flux = Bπa²/∆t

Using the above equations, we get

E2πb = Bπa²/∆t

QE is the electric force experienced by the ring

Torque on the ring is Q.Ba²/2∆t

Change in angular momentum = torque × ∆t

The initial momentum is zero and the final momentum is Q.Ba²/2

ꞷ = Q.Ba²/2mb²

Q6.28Written answer

A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle θ with respect to the horizontal. The circuit is closed through a perfect conductor at the top. There is a constant magnetic field B along the vertical direction. If the rod is initially at rest, find the velocity of the rod as a function of time.

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Worked solution

The angle between B and PQ = 90^o

dϕ = B.dA

dϕ = B v d cos θ

-ε = B v d cos θ

I = -Bvd/R cos θ

Solving the above equation using Newton’s second law, we get, v as

v = α g sin θ [1 – e ^t/ꞷ]

Q6.29Written answer

Find the current in the sliding rod AB (resistance = R). B is constant and is out of the paper. Parallel wires have no resistance. v is constant. Switch S is closed at time t = 0.

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Worked solution

The current induced in the loop is It = ε/R

It = 1/R.d/dt BA

It = vBd/R

The angle between B and A is zero.

As the switch S is closed at t = o,

Charge through the capacitor, is Q(t) = Cv

Current through the capacitor, I(c) = Q(t)/RC

Using the above information, we can calculate the current through the circuit as I = BvdC/RC e ^–t/RC.

Q6.30Written answer

Find the current in the sliding rod AB (resistance = R). B is constant and is out of the paper. Parallel wires have no resistance. v is constant. Switch S is closed at time t = 0.

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Worked solution

The angle between A and B = 0^o

Therefore, the emf is Bvd.

-LdI(t)/dt + Bvd = IR

LdI(t)/dt + RI (t) = Bvd

Solving the equation, we get

I = Bvd/R [1-e^-Rt/2]

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