JEE Challenger
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PhyHardAdv 2023 • Paper 2

Speed of Moving Rectangular Loop Near Current Carrying Wire

A rectangular conducting loop of length 4 cm4\text{ cm} and width 2 cm2\text{ cm} is in the xyxy-plane, as shown in the figure. It is being moved away from a thin and long conducting wire along the direction 32x^+12y^\frac{\sqrt{3}}{2}\hat{x} + \frac{1}{2}\hat{y} with a constant speed vv. The wire is carrying a steady current I=10 AI = 10\text{ A} in the positive xx-direction. A current of 10 μA10\text{ }\mu\text{A} flows through the loop when it is at a distance d=4 cmd = 4\text{ cm} from the wire. If the resistance of the loop is 0.1 Ω0.1\text{ }\Omega, then the value of vv is ______ m s−1\text{m s}^{-1}.

[Given: The permeability of free space μ0=4π×10−7 N A−2\mu_0 = 4\pi \times 10^{-7}\text{ N A}^{-2}]

Question Diagram 1
Official Numerical Answer4

Topics & Concepts

Electromagnetic InductionFaraday's LawLenz's LawBiot-Savart Law

Step-by-Step Solution

To find the speed vv of the rectangular loop, we use Faraday's Law of Electromagnetic Induction.

1. Magnetic Field and Magnetic Flux

The magnetic field B⃗\vec{B} at a distance yy from a long straight wire carrying a current II along the positive xx-direction is directed perpendicular to the xyxy-plane (along the +z^+\hat{z} direction) and has a magnitude given by Biot-Savart Law: B(y)=μ0I2πyB(y) = \frac{\mu_0 I}{2\pi y}

Let the position of the bottom edge of the loop be y0y_0 at time tt. The dimensions of the loop are:

  • Width along xx-axis, a=2 cm=0.02 ma = 2\text{ cm} = 0.02\text{ m}
  • Length along yy-axis, b=4 cm=0.04 mb = 4\text{ cm} = 0.04\text{ m}

The magnetic flux Φ\Phi through the rectangular loop is given by: Φ=∫y0y0+bB(y)⋅(a dy)=μ0Ia2π∫y0y0+bdyy=μ0Ia2πln⁡(y0+by0)\Phi = \int_{y_0}^{y_0 + b} B(y) \cdot (a \, dy) = \frac{\mu_0 I a}{2\pi} \int_{y_0}^{y_0 + b} \frac{dy}{y} = \frac{\mu_0 I a}{2\pi} \ln\left( \frac{y_0 + b}{y_0} \right)

2. Induced Electromotive Force (EMF)

The loop moves with velocity: v⃗=vxx^+vyy^=v(32x^+12y^)\vec{v} = v_x \hat{x} + v_y \hat{y} = v \left( \frac{\sqrt{3}}{2}\hat{x} + \frac{1}{2}\hat{y} \right)

Since the magnetic field depends only on yy, motion along the xx-axis (vxv_x) does not contribute to a change in magnetic flux. Only the component of velocity along the yy-axis, vy=v2v_y = \frac{v}{2}, changes the position y0y_0 of the loop.

Using Faraday's law of induction, the magnitude of the induced electromotive force ∣E∣|\mathcal{E}| is: ∣E∣=∣dΦdt∣=∣dΦdy0dy0dt∣|\mathcal{E}| = \left| \frac{d\Phi}{dt} \right| = \left| \frac{d\Phi}{dy_0} \frac{dy_0}{dt} \right|

Differentiating Φ\Phi with respect to y0y_0: dΦdy0=μ0Ia2π(1y0+b−1y0)=−μ0Iab2πy0(y0+b)\frac{d\Phi}{dy_0} = \frac{\mu_0 I a}{2\pi} \left( \frac{1}{y_0 + b} - \frac{1}{y_0} \right) = -\frac{\mu_0 I a b}{2\pi y_0 (y_0 + b)}

Therefore, since vy=dy0dt=v2v_y = \frac{dy_0}{dt} = \frac{v}{2}: ∣E∣=μ0Iab2πy0(y0+b)vy=μ0Iab2πy0(y0+b)⋅v2|\mathcal{E}| = \frac{\mu_0 I a b}{2\pi y_0 (y_0 + b)} v_y = \frac{\mu_0 I a b}{2\pi y_0 (y_0 + b)} \cdot \frac{v}{2}

3. Calculation of Induced Current and EMF

From Ohm's law, the induced current ii is related to the resistance RR of the loop by: ∣E∣=iR|\mathcal{E}| = i R

Given data:

  • Current I=10 AI = 10\text{ A}
  • Permeability μ0=4π×10−7 N A−2\mu_0 = 4\pi \times 10^{-7}\text{ N A}^{-2}
  • a=0.02 ma = 0.02\text{ m}
  • b=0.04 mb = 0.04\text{ m}
  • Distance y0=d=0.04 my_0 = d = 0.04\text{ m}
  • Induced current i=10 μA=10−5 Ai = 10\ \mu\text{A} = 10^{-5}\text{ A}
  • Resistance R=0.1 ΩR = 0.1\ \Omega

The magnitude of induced EMF is: ∣E∣=10−5 A×0.1 Ω=10−6 V|\mathcal{E}| = 10^{-5}\text{ A} \times 0.1\ \Omega = 10^{-6}\text{ V}

Substitute the numerical values into the formula for ∣E∣|\mathcal{E}|: 10−6=(4π×10−7)×10×0.02×0.042π×0.04×(0.04+0.04)×vy10^{-6} = \frac{(4\pi \times 10^{-7}) \times 10 \times 0.02 \times 0.04}{2\pi \times 0.04 \times (0.04 + 0.04)} \times v_y

Simplify the factor: μ0I2π=2×10−7 T⋅m A−1\frac{\mu_0 I}{2\pi} = 2 \times 10^{-7}\text{ T}\cdot\text{m A}^{-1}

aby0(y0+b)=0.02×0.040.04×0.08=0.020.08=14\frac{a b}{y_0 (y_0 + b)} = \frac{0.02 \times 0.04}{0.04 \times 0.08} = \frac{0.02}{0.08} = \frac{1}{4}

So: 10−6=(2×10−7)×10×14×vy10^{-6} = \left( 2 \times 10^{-7} \right) \times 10 \times \frac{1}{4} \times v_y

10−6=0.5×10−6vy10^{-6} = 0.5 \times 10^{-6} v_y

vy=2 m s−1v_y = 2\text{ m s}^{-1}

Since vy=v2v_y = \frac{v}{2}: v2=2  ⟹  v=4 m s−1\frac{v}{2} = 2 \implies v = 4\text{ m s}^{-1}

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