A wire cd of length l and mass m is sliding w | vedclass

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(B) As the wire $cd$ moves with velocity $v$ in a magnetic field $B$,an induced electromotive force $(EMF)$ is generated across the wire given by $\va

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) As the wire $cd$ moves with velocity $v$ in a magnetic field $B$,an induced electromotive force $(EMF)$ is generated across the wire given by $\varepsilon = Blv$.Since the wire is part of a closed circuit with resistance $R$,an induced current $I$ flows through it,given by $I = \frac{\varepsilon}{R} = \frac{Blv}{R}$.The wire experiences a magnetic force $F_m$ acting upwards,given by $F_m = IlB = \left( \frac{Blv}{R} \right) lB = \frac{B^2l^2v}{R}$.For the wire to move with a constant velocity,the net force acting on it must be zero. Therefore,the upward magnetic force must balance the downward gravitational force $(mg)$:$F_m = mg$$\frac{B^2l^2v}{R} = mg$Solving for $v$,we get $v = \frac{mgR}{B^2l^2}$.

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