XPQY is a vertical smooth long loop having a total resistance R where… | JEE Main 2026 (22 Jan, Shift 1)

Q10JEE Main 2026 · 22 JAN · SHIFT 1Medium

Motional EMF

$XPQY$ 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 $\underline{\qquad}\ \text{m/s. }(g =$ acceleration due to gravity)

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Answer: (a)

Only the part of rod between the rails (length $l$ ) is in the circuit, so $e = Blv$ .
$I = \dfrac{Blv}{R}$ , magnetic force $F = BIl = \dfrac{B^2l^2v}{R}$ upward.
At terminal speed $F = mg$ : $v_t = \dfrac{mgR}{B^2l^2}$
So the answer is (a) $\dfrac{mgR}{B^2l^2}$ .

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