JEE Advanced 2023 Paper 1 · Q17 · EM Induction+LR

Question

A thin conducting rod $MN$ of mass $20$ gm, length $25$ cm and resistance $10\,\Omega$ is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field $B_0 = 4$ T directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time $t = 0$ and it moves down along the rails. Assume air drag is negligible. Match each quantity in List-I with an appropriate value from List-II. [Given: $g = 10$ m·s$^{-2}$ and $e^{-1} = 0.4$]
List-I:
(P) At $t = 0.2$ s, the magnitude of the induced emf in Volt
(Q) At $t = 0.2$ s, the magnitude of the magnetic force in Newton
(R) At $t = 0.2$ s, the power dissipated as heat in Watt
(S) The magnitude of terminal velocity of the rod in m·s$^{-1}$
List-II: (1) $0.07$, (2) $0.14$, (3) $1.20$, (4) $0.12$, (5) $2.00$

SolveKar AI · Accepted Answer

Equation of motion: $m\dot{v} = mg - B^2\ell^2 v/R$. With $m = 0.02$ kg, $B\ell = 4\cdot 0.25 = 1$, $R = 10\,\Omega$: terminal velocity $v_T = mgR/(B\ell)^2 = (0.02)(10)(10)/1 = 2$ m/s — (S)→(5). Time constant $	au = mR/(B\ell)^2 = 0.2$ s. At $t = 0.2$ s ($= 	au$): $v = v_T(1 - e^{-1}) = 2(1-0.4) = 1.2$ m/s. (P) emf = $B\ell v = 1\cdot 1.2 = 1.20$ V — (3). $i = $emf$/R = 0.12$ A. (Q) Magnetic force = $i B\ell = 0.12$ N — (4). (R) Heat $= i^2 R = 0.144$ W $\approx 0.14$ — (2).

JEE Advanced 2023 Paper 1 · Q17 · EM Induction+LR

Solution