JEE Advanced 2021 Paper 1 Q15 Physics Electrostatics & Circuits EM Induction+Circuits Hard

JEE Advanced 2021 Paper 1 · Q15 · EM Induction+Circuits

A long straight wire carries a current $I = 2\,\text{A}$. A semi-circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both rails are parallel to the wire. The wire, the rod, and the rails all lie in the same horizontal plane. The two ends of the semi-circular rod are at distances $1\,\text{cm}$ and $4\,\text{cm}$ from the wire (so the rod spans the gap between the two rails, which are at $1\,\text{cm}$ and $4\,\text{cm}$ from the wire respectively). At time $t = 0$, the rod starts moving along the rails (away from the current-carrying wire) with a constant speed $v = 3.0\,\text{m/s}$. A resistor $R = 1.4\,\Omega$ and a capacitor $C_0 = 5.0\,\mu\text{F}$ are connected in series between the rails, closing the circuit. At $t = 0$, $C_0$ is uncharged. Which of the following statement(s) is(are) correct? [$\mu_0 = 4\pi \times 10^{-7}$ SI units; take $\ln 2 = 0.7$]

  1. A. Maximum current through R is 1.2 × 10⁻⁶ ampere
  2. B. Maximum current through R is 3.8 × 10⁻⁶ ampere
  3. C. Maximum charge on capacitor C₀ is 8.4 × 10⁻¹² coulomb
  4. D. Maximum charge on capacitor C₀ is 2.4 × 10⁻¹² coulomb
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C

Solution

EMF induced in the rod = ∫₁⁴ (μ₀I/2πx)·v dx = (μ₀Iv/2π)·ln(4) = (μ₀Iv/π)·ln 2. Since v is constant, ε is constant after t = 0, so the RC transient is from 0 to V_∞ = ε. Maximum current = ε/R (at t = 0⁺): i_max = (μ₀Iv/πR)·ln 2 = (4π×10⁻⁷ × 2 × 3 × 0.7)/(π × 1.4) = 1.2 × 10⁻⁶ A ⇒ (A) correct. Max charge = C₀·ε = 5×10⁻⁶ × (4π×10⁻⁷·2·3·0.7/π) = 5×10⁻⁶ × 1.68×10⁻⁶ = 8.4×10⁻¹² C ⇒ (C) correct. Units: ε has V (= T·m·m/s = V), I=A, q=C — all dimensionally consistent.

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