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Physics · Electromagnetic Induction

JEE Advanced 2021 — Paper 1 — Question 7

A long straight wire carries a current, I=2\mathrm{I}=2 ampere. A semi-circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both the rails are parallel to the wire. The wire, the rod and the rails lie in the same horizontal plane, as shown in the figure.

Two ends of the semi-circular rod are at distances 1 cm and 4 cm from the wire. At time t=0t=0, the rod starts moving on the rails with a speed

v=3.0 m/s\mathrm{v}=3.0 \mathrm{~m} / \mathrm{s} (see the figure) A resistor R=1.4Ω\mathrm{R}=1.4 \Omega and a capacitor C0=5.0μ F\mathrm{C}_{0}=5.0 \mu \mathrm{~F}

are connected in series between the rails. At time t=0,C0\mathrm{t}=0, \mathrm{C}_{0} is uncharged.

Which of the following statement(s) is(are) correct? [ μ0=4π×10−7\mu_{0}=4 \pi \times 10^{-7} SI units. Take ℓn2=0.7\ell \mathrm{n} 2=0.7 ]

Question figure
  1. Option A:

    Maximum current through R is 1.2×10−61.2 \times 10^{-6} ampere

    Correct
  2. Option B:

    Maximum current through R is 3.8×10−63.8 \times 10^{-6} ampere

  3. Option C:

    Maximum charge on capacitor C0\mathrm{C}_{0} is 8.4×10−128.4 \times 10^{-12} coulomb

    Correct
  4. Option D:

    Maximum charge on capacitor C0\mathrm{C}_{0} is 2.4×10−122.4 \times 10^{-12} coulomb

Answer: A, C

Step-by-step solution

Emf induced across the semi-circular conducting rod. ε=∫14μ0Ivdx⁡2πx=μ0Iv⁡2πln⁡(4)=μ0Ivπln⁡(2)\varepsilon=\int_{1}^{4} \frac{\mu_{0} \operatorname{Ivdx}}{2 \pi x}=\frac{\mu_{0} \operatorname{Iv}}{2 \pi} \ln (4)=\frac{\mu_{0} \mathrm{Iv}}{\pi} \ln (2)

Since the semi-circular conducting rod is moving with a constant speed

v=3 m/sv=3 \mathrm{~m} / \mathrm{s}, then ε=μ0Ivπln⁡(2)=\varepsilon=\frac{\mu_{0} \mathrm{Iv}}{\pi} \ln (2)=

constant Maximum current through the resistor R is

imax⁡=εR=μ0IvπRln⁡(2)=4×10−7×2×3×0.71.4=1.2×10−6\mathrm{i}_{\max }=\frac{\varepsilon}{\mathrm{R}}=\frac{\mu_{0} \mathrm{Iv}}{\pi \mathrm{R}} \ln (2)=\frac{4 \times 10^{-7} \times 2 \times 3 \times 0.7}{1.4}=1.2 \times 10^{-6} ampere. Maximum charge on the capacitor

C0\mathrm{C}_{0} is

qmax⁡=C0ε=C0(μ0Ivπln⁡(2))=5×10−6×4×10−7×2×3×0.7=8.4×10−12\mathrm{q}_{\max }=\mathrm{C}_{0} \varepsilon=\mathrm{C}_{0}\left(\frac{\mu_{0} \mathrm{Iv}}{\pi} \ln (2)\right)=5 \times 10^{-6} \times 4 \times 10^{-7} \times 2 \times 3 \times 0.7=8.4 \times 10^{-12} coulomb.

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2021
Paper
Paper 1
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Motional EMF