A copper rod of mass m slides under gravity on two smooth parallel… | JEE Main 2018 (15 Apr, Shift 2)

Q49JEE Main 2018 · 15 APR · SHIFT 2Difficult

Motional EMF

A copper rod of mass m slides under gravity on two smooth parallel rails, with separation l and set at an angle of $\theta$ with the horizontal. At the bottom rails are joined by a resistance R. There is a uniform magnetic field B normal to the plane of the rails, as shown in the igure. The terminal speed of the copper rod is:

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Answer: (c)

Gravity component along the rails: $mg\sin\theta$ .
$\vec{B}$ is normal to the rails' plane, so $e = Blv$ , $I = \dfrac{Blv}{R}$ , and drag force $F = BIl = \dfrac{B^2l^2v}{R}$ up the incline.
Terminal speed: $\dfrac{B^2l^2v}{R} = mg\sin\theta$
So the answer is (c) $v = \dfrac{mgR\sin\theta}{B^2l^2}$ .

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