A horizontal wire is free to slide on the vertical rails of a conducting frame as shown in figure. The wire has a mass m and length l and the resistance of the circuit is R. If a uniform magnetic field B is directed perpendicular to the frame, the terminal speed of the wire as it falls under the force of gravity is

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$\frac{mgR}{BI}$

$\frac{mgl}{BR}$

$\frac{B^{2} l^{2}}{mgR}$

$\frac{mgR}{B^{2} l^{2}}$

answer is D.

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Detailed Solution

Given data: As depicted in the figure, a horizontal wire can freely glide along vertical rails of a conducting frame. The resistance of the circuit is R, and the wire has a mass of m and a length of l.

Detailed solution:

$force, F = IlB . 0=IlB-mg mg = \frac{ε}{R} lB V=IR \frac{ε}{R} = I . \frac{ε}{R} = \frac{{BlV}_{T}}{R} \frac{{BlV}_{T}}{R} lB = mg The force of gravity, V_{T} = \frac{mgR .}{B^{2} l^{2}}$