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QJEEAdv21Ph6308: JEE Advanced 2021 Physics - Class 12 JEEAdv21Ph6308 Solution: A long straight wire carries a current, I = 2 ampere. A…

A long straight wire carries a current, 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 the distances 1 cm and 4 cm from the wire. At time t = 0, the rod starts moving on the rails with a speed v = 3.0 m/s (see the figure).

A resistor R = 1.4 Ω\Omega and a capacitor C0 = 5.0 μ\mu F are connected in series between the rails. At time t = 0, C0 is uncharged. Which of the following statement(s) is(are) correct? [ μ\mu 0 = 4 π\pi ×\times 10 −- 7 SI units. Take ln 2 = 0.7]

JEE Advanced 2021 Paper 1 Online Physics - Magnetism Question 33 English

Multiple Choice Options

  • A Maximum current through R is 1.2 ×\times 10 −- 6 ampere
  • B Maximum current through R is 3.8 ×\times 10 −- 6 ampere
  • C Maximum charge on capacitor C0 is 8.4 ×\times 10 −- 12 coulomb
  • D Maximum charge on capacitor C0 is 2.4 ×\times 10 −- 12 coulomb

Step-by-Step Verified Solution

JEE Advanced 2021 Paper 1 Online Physics - Magnetism Question 33 English Explanation
Let a small element of length dxd x be chosen at a distance xx from the wire. Induced emf de across it is given by

dEamp;=Bv(dx)amp;=μ0I2πxv(dx)EPQamp;=μ0Iv2π∫14dxxEPQamp;=μ0Iv2πln⁡4=μ0Iv2π×2ln⁡2\begin{aligned}d \mathrm{E} & =\mathrm{B} v(d x) \\\\& =\frac{\mu_0 \mathrm{I}}{2 \pi x} v(d x) \\\\\mathrm{E}_{\mathrm{PQ}} & =\frac{\mu_0 \mathrm{I} v}{2 \pi} \int_1^4 \frac{d x}{x} \\\\\mathrm{E}_{\mathrm{PQ}} & =\frac{\mu_0 \mathrm{I} v}{2 \pi} \ln 4=\frac{\mu_0 \mathrm{I} v}{2 \pi} \times 2 \ln 2\end{aligned}

On substituting all the values in above, we get

EPQamp;=2×10−7×2×3×2×0.7amp;=16.8×10−7 V\begin{aligned}\mathrm{E}_{\mathrm{PQ}} & =2 \times 10^{-7} \times 2 \times 3 \times 2 \times 0.7 \\\\& =16.8 \times 10^{-7} \mathrm{~V}\end{aligned}

At t=0,Ct=0, \mathrm{C} acts as a short. So, maximum current flows through RR .

imax⁡amp;=EPQR=16.8×10−71.4amp;=12×10−7 Aamp;=1.2×10−6 A\begin{aligned}i_{\max } & =\frac{\mathrm{E}_{\mathrm{PQ}}}{\mathrm{R}}=\frac{16.8 \times 10^{-7}}{1.4} \\\\& =12 \times 10^{-7} \mathrm{~A} \\\\& =1.2 \times 10^{-6} \mathrm{~A}\end{aligned}

At t→∞,Ct \rightarrow \infty, C acts as on open. So,

qmax⁡amp;=CEPQ=5×10−6×16.8×10−7Camp;=84×10−13Camp;=8.4×10−12C\begin{aligned}q_{\max } & =\mathrm{CE}_{\mathrm{PQ}}=5 \times 10^{-6} \times 16.8 \times 10^{-7} \mathrm{C} \\\\& =84 \times 10^{-13} \mathrm{C} \\\\& =8.4 \times 10^{-12} \mathrm{C}\end{aligned}

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