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# A horizontal wire AB of length l is free to slide on a vertical conducting frame as shown. The wire has mass m and length l and resistance of circuit is R. The system is placed in a uniform magnetic field B directed perpendicular to the frame. The wire after falling a distance under the influence of gravity aquires a terminal speed equal to

$(a)\;\large\frac{B^2l^2}{mgR} \\ (b)\;\large\frac{mgR}{b^2l^2} \\(c)\;\large\frac{mgR^2}{B^2l} \\(d)\;\large\frac{mgl}{BR}$

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When the falling wire aquires a terminal velocity the net force on the wire must be zero, Therefore the weight of the body must balance the force experienced by the wire in magnetic field.
If $V_{T}$ is the terminal velocity of the wire the included emf e is
$e= Bv_{T}e$
and current $i=\large\frac{e}{R}=\frac{Bv_Tl}{R}$
Force experienced $=ilB$
$\therefore ilB=mg$
$\bigg(\large\frac{Bv_T l}{R}\bigg)$$lB=mg$
$v_T=\large\frac{mgR}{B^2l^2}$
Hence b is the correct answer.

answered Mar 21, 2014 by
edited Sep 24, 2014