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← JEE Advanced 2023 · Paper 2
PhysicsInteger+4Question 26 of 51

JEE Advanced 2023 · Paper 2, Physics question 9

A rectangular conducting loop of length 4 cm and width 2 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^\dfrac{\sqrt3}{2}\hat x + \dfrac12\hat y with a constant speed v. The wire is carrying a steady current I=10I = 10 A in the positive xx-direction. A current of 10 μ10\ \muA flows through the loop when it is at a distance d=4d = 4 cm from the wire. If the resistance of the loop is 0.1 Ω0.1\ \Omega, then the value of v is _____ m s−1\mathrm{m\,s^{-1}}.

[Given: The permeability of free space μ0=4π×10−7 N A−2\mu_0 = 4\pi\times10^{-7}\ \mathrm{N\,A^{-2}}]

A long straight wire along the x-axis carries I = 10 A in the +x direction. Above it, a rectangular loop 2 cm wide (along x) and 4 cm long (along y), with its lower edge at distance d from the wire. A dashed arrow v from the loop's centre points up and to the right (at 30 degrees to x). Axes: y up, x to the right.
Answer and worked solution

Official answer: 4

Flux. The wire is along xx. The loop has width w=2w=2 cm along xx and length L=4L=4 cm along yy, with its near side at distance dd:
Φ=μ0Iw2πln⁡d+Ld.\Phi=\frac{\mu_0Iw}{2\pi}\ln\frac{d+L}{d}.

Rate of change. Only the velocity component away from the wire changes Φ\Phi: d˙=v2\dot d=\frac v2.
∣E∣=μ0Iw2π⋅Ld(d+L)⋅v2.|\mathcal E|=\frac{\mu_0Iw}{2\pi}\cdot\frac{L}{d(d+L)}\cdot\frac v2.

Numbers.

  • μ0I2πw=2×10−6×0.02=4×10−8\frac{\mu_0I}{2\pi}w=2\times10^{-6}\times0.02=4\times10^{-8};
  • Ld(d+L)=0.040.04×0.08=12.5 m−1\frac{L}{d(d+L)}=\frac{0.04}{0.04\times0.08}=12.5\ \text{m}^{-1}.

So E=2.5×10−7 v\mathcal E=2.5\times10^{-7}\,v volts.

Current.
i=ER=2.5×10−6 v=10×10−6  ⟹  v=4 m/s.i=\frac{\mathcal E}{R}=2.5\times10^{-6}\,v=10\times10^{-6}\implies v=4\ \text{m/s}.

Topics:Faraday's and Lenz's lawsBiot-Savart law
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