JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Electromagnetic Induction

XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is

l

. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L (

L > l

) and mass

m

is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is ______ m/s.
(

g =

acceleration due to gravity)

Answer: (D)

Solution

As the rod falls with speed

v

, only the length

l

between the rails lies in the circuit, so the induced emf is

ε = B v l

and the current is

i = \frac{B v l}{R}

. The magnetic force on the rod,

F = B i l = \frac{B ^{2} l ^{2} v}{R}

, acts upward and grows with

v

. At the terminal speed this balances gravity,

m g = \frac{B ^{2} l ^{2} v _{t}}{R}

, so

v_{t} = \frac{m g R}{B ^{2} l ^{2}} .

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option