JEE Advanced 2025 Paper 1 Q05 Physics Electrostatics & Circuits EM Induction Hard

JEE Advanced 2025 Paper 1 · Q05 · EM Induction

A conducting square loop of side $L$, mass $M$, and resistance $R$ moves in the XY plane with edges parallel to the X and Y axes. The region $y \ge 0$ has a uniform magnetic field $\vec{B} = B_0\,\hat{k}$ (out of the XY plane in the $+Z$ direction); $B$ is zero everywhere else. At $t = 0$ the loop, lying entirely in $y < 0$, starts to enter the magnetic field with initial velocity $v_0\,\hat{j}$ (in $+Y$ direction). Define $K = B_0^2 L^2/(R M)$. Ignore self-inductance and gravity. Which statements are correct?

  1. A. If $v_0 = 1.5\,K L$, the loop stops before it enters completely into the magnetic-field region.
  2. B. When the complete loop is inside the magnetic-field region, the net force on the loop is zero.
  3. C. If $v_0 = K L/10$, the loop comes to rest at $t = (1/K)\ln(5/2)$.
  4. D. If $v_0 = 3\,K L$, the complete loop enters the field at time $t = (1/K)\ln(3/2)$.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:B, D

Solution

While entering, the top edge of length $L$ is in the field; flux $\Phi = B_0 L y$, EMF $= B_0 L (dy/dt)$, induced current $i = B_0 L v/R$, retarding force $F = -B_0^2 L^2 v/R = -M K v$. So $dv/dt = -K v \implies v(t) = v_0\,e^{-K t}$, and $y(t) = (v_0/K)(1 - e^{-K t})$. The loop fully enters when $y = L$, so the time is $t_1 = (1/K)\ln\!\bigl(v_0/(v_0 - K L)\bigr)$, provided $v_0 > K L$. (A) $v_0 = 1.5\,K L \implies y_{\max} = v_0/K = 1.5L > L$: loop DOES enter completely. (A) FALSE. (B) Once fully inside the uniform field, flux is constant $\implies$ no EMF, no induced current, no magnetic force, and gravity is ignored. Net force $= 0$. (B) TRUE. (C) $v_0 = K L/10 < K L$: $y_{\max} = v_0/K = L/10 < L$. The loop never fully enters; it asymptotically slows but comes to rest only as $t \to \infty$ (exponential decay). The expression $t = (1/K)\ln(5/2)$ is finite, hence incorrect. (C) FALSE. (D) $v_0 = 3 K L \implies t_1 = (1/K)\ln(3KL/(3KL - KL)) = (1/K)\ln(3/2)$. (D) TRUE. Answer: B, D.

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