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JEE Advanced 2021Paper 1
Exam
JEE Advanced
Year
2021
Paper
Paper 1
Subject
Physics
Chapter
Electromagnetic Induction

JEE Advanced 2021 · Paper 1 · Physics · Electromagnetic Induction

A long straight wire carries a current, I = 2 ampere. A...

A long straight wire carries a current, I=2I = 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 1cm1cm and 4cm4cm from the wire. At time t=0t = 0, the rod starts moving on the rails with a speed v=3.0ms−1v = 3 . 0m{s}^{- 1} (see the figure).

A resistor R=1.4ΩR = 1.4Ω and a capacitor C0=5.0μF{C}_{0} = 5.0\mu F are connected in series between the rails. At time t=0t = 0, C0{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 ln⁡2=0.7\ln 2 = 0.7]

A long straight wire carries a current, I = 2 ampere. A semi-circular conducting rod is placed beside it on two

Options

  1. AMaximum current through RR is 1.2×10−61.2 \times {10}^{- 6} ampere
  2. BMaximum current through RR is 3.8×10−63.8 \times {10}^{- 6} ampere.
  3. CMaximum charge on capacitor C0{C}_{0} is 8.4×10−128.4 \times {10}^{- 12} coulomb.
  4. DMaximum charge on capacitor C0{C}_{0} is 2.4×10−122 .4 \times {10}^{- 12} coulomb.

Correct answer

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

Step-by-step solution

Magnetic field due the long wire, B=μ0I2\pixB = \dfrac{{\mu}_{0}I}{2\pix}

Motional emf in small element,

dε=Bvdx=3μ0(2)2\pixdxdε = Bv d x = \dfrac{3{\mu}_{0}\left(2\right)}{2\pix}dx

⇒ε=3μ0π∫141xdx⇒ ε = \dfrac{3{\mu}_{0}}{\pi}{∫}_{1}^{4}\dfrac{1}{x}dx

∴ε=3μ0πln⁡4∴ ε = \dfrac{3{\mu}_{0}}{\pi}\ln 4=16.8×10−7= 16.8 \times {10}^{- 7}

Maximum charge on capacitor,

{q}_\max = εC

=(16.8×10−7)(5×10−6)= \left(16.8 \times {10}^{- 7}\right)\left(5 \times {10}^{- 6}\right)

=84×10−13C=8.4×10−12C= 84 \times {10}^{- 13}C = 8.4 \times {10}^{- 12}C

Maximum current in circuit,

{I}_\max = \dfrac{ε}{R} = \dfrac{16.8 \times {10}^{- 7}}{1.4}

=1.2×10−6A= 1 . 2 \times {10}^{- 6}A

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