JEE Main 2026 — Jan 22, Shift 1 — Physics solution

JEE Main2026Jan 22, Shift 1Physics

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)

A
$\frac{2 m g R}{B ^{2} l ^{2}}$
B
$\frac{8 m g R}{B ^{2} l ^{2}}$
C
$\frac{2 m g R}{B ^{2} L ^{2}}$
D
$\frac{m g R}{B ^{2} l ^{2}}$

Solution

The induced EMF depends only on the portion of the rod between the two rails, of length $l$ (the part outside the rails doesn’t enclose any flux change). So $ε = B v l$ , and the induced current is $i = \frac{B v l}{R}$ .

The magnetic force on the conducting segment is $F_{ma g} = B i l = \frac{B ^{2} l ^{2} v}{R}$ , acting upward to oppose motion.

At terminal speed, gravity balances this magnetic drag: $m g = \frac{B ^{2} l ^{2} v _{T}}{R}$ , so $v_{T} = \frac{m g R}{B ^{2} l ^{2}}$ .

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