JEE Challenger
More from Electromagnetic Induction
PhyHardAdv 2019 • Paper 2

Current in Inductor Circuit with Moving Conducting Wire

A 10 cm10\text{ cm} long perfectly conducting wire PQPQ moves with a constant speed of 1 cm/s1\text{ cm/s} along a pair of horizontal rails with zero resistance. One end of the rails is connected to an inductor L=1 mHL = 1\text{ mH} and a resistor R=1 ΩR = 1\ \Omega as illustrated in the figure. The horizontal rails, LL, and RR lie in the same plane inside a uniform magnetic field B=1 TB = 1\text{ T} directed perpendicular into the plane. When the switch SS is closed at a given moment, the current in the circuit after 1 ms1\text{ ms} is x×10−3 Ax \times 10^{-3}\text{ A}, where the value of xx is _____.

[Assume that the speed of the wire PQPQ remains constant (1 cm/s1\text{ cm/s}) after switch SS is closed. Given: e−1=0.37e^{-1} = 0.37, where ee is the base of natural logarithm]

Question Diagram 1
Official Numerical Answer0.63

Topics & Concepts

Electromagnetic InductionFaraday's LawLenz's LawSelf and Mutual Inductance

Step-by-Step Solution

When the conducting wire PQPQ moves with a constant velocity vv perpendicular to the uniform magnetic field BB, a motional electromotive force (emf) E\mathcal{E} is induced across its length ll.

The magnitude of the induced emf E\mathcal{E} is given by: E=Bvl\mathcal{E} = B v l

Given:

  • Magnetic field, B=1 TB = 1\text{ T}
  • Velocity of wire, v=1 cm/s=10−2 m/sv = 1\text{ cm/s} = 10^{-2}\text{ m/s}
  • Length of wire, l=10 cm=0.1 ml = 10\text{ cm} = 0.1\text{ m}

Substituting the given values: E=1×10−2×0.1=10−3 V\mathcal{E} = 1 \times 10^{-2} \times 0.1 = 10^{-3}\text{ V}

By Fleming's Right-Hand Rule (or using v⃗×B⃗\vec{v} \times \vec{B}), the induced emf acts such that point PP is at a higher potential than point QQ. When the switch SS is closed at t=0t = 0, the circuit acts as an LRLR circuit connected across a constant DC voltage source of emf E\mathcal{E}.

The differential equation governing the growth of current I(t)I(t) in the LRLR circuit is: E−LdIdt−IR=0\mathcal{E} - L \frac{dI}{dt} - I R = 0

Solving this differential equation with the initial condition I(0)=0I(0) = 0, we get the current as a function of time tt: I(t)=ER(1−e−Rt/L)I(t) = \frac{\mathcal{E}}{R} \left( 1 - e^{-Rt/L} \right)

Given parameters:

  • Resistance, R=1 ΩR = 1\ \Omega
  • Inductance, L=1 mH=10−3 HL = 1\text{ mH} = 10^{-3}\text{ H}
  • Time, t=1 ms=10−3 st = 1\text{ ms} = 10^{-3}\text{ s}

The time constant τ\tau of the circuit is: τ=LR=10−3 H1 Ω=10−3 s=1 ms\tau = \frac{L}{R} = \frac{10^{-3}\text{ H}}{1\ \Omega} = 10^{-3}\text{ s} = 1\text{ ms}

Since t=1 ms=τt = 1\text{ ms} = \tau, the exponent simplifies to: RtL=1×10−310−3=1\frac{R t}{L} = \frac{1 \times 10^{-3}}{10^{-3}} = 1

Substituting t=1 mst = 1\text{ ms} into the current equation: I(1 ms)=10−31(1−e−1) AI(1\text{ ms}) = \frac{10^{-3}}{1} \left( 1 - e^{-1} \right)\text{ A}

Using the given value e−1=0.37e^{-1} = 0.37: I(1 ms)=10−3×(1−0.37) A=0.63×10−3 AI(1\text{ ms}) = 10^{-3} \times (1 - 0.37)\text{ A} = 0.63 \times 10^{-3}\text{ A}

Comparing this with I=x×10−3 AI = x \times 10^{-3}\text{ A}, we obtain: x=0.63x = 0.63

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