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MHT CET 2023 (14 May, Shift 1) · Other: Induction & Displacement Current

Other: Induction & Displacement Current question from MHT CET 2023 (14 May, Shift 1)

Q9MHT CET 2023 · 14 MAY · SHIFT 1Medium

Other: Induction & Displacement Current

A horizontal wire of mass ' $m$ ', length ' $l$ ' and resistance ' $R$ ' is sliding on the vertical rails on which uniform magnetic field ' $B$ ' is directed perpendicular. The terminal speed of the wire as it falls under the force of gravity is ( $g =$ acceleration due to gravity)

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Answer: (d)

Falling wire: emf $= Blv$ , current $I = \frac{Blv}{R}$ , upward magnetic force $F = BIl = \frac{B^2 l^2 v}{R}$ .
At terminal speed the force balances weight: $\frac{B^2 l^2 v_t}{R} = mg$
$v_t = \frac{mgR}{B^2 l^2}$
So the answer is (d).

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