A metal rod of length 'L' completes the circu | vedclass

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Question

(B) When a metal rod of length '$L$' moves with a constant velocity '$V$' in a magnetic field '$B$',an induced electromotive force $(EMF)$ is generate

Vedclass · Accepted Answer

$\frac{B^2 L^2 V}{R}$

Vedclass · Answer

(B) When a metal rod of length '$L$' moves with a constant velocity '$V$' in a magnetic field '$B$',an induced electromotive force $(EMF)$ is generated across the rod, given by $\varepsilon = BLV$.Since the rod completes a circuit with total resistance '$R$',the induced current '$I$' flowing through the circuit is $I = \frac{\varepsilon}{R} = \frac{BLV}{R}$.The magnetic force '$F_m$' acting on the current-carrying rod in the magnetic field is given by $F_m = I L B$.Substituting the value of '$I$',we get $F_m = (\frac{BLV}{R}) L B = \frac{B^2 L^2 V}{R}$.To move the rod at a constant speed, an external force '$F$' must be applied equal and opposite to the magnetic force. Therefore, the required force is $F = \frac{B^2 L^2 V}{R}$.

$A$ metal rod of length '$L$' completes the circuit as shown. The area of the circuit is perpendicular to the magnetic field '$B$'. The total resistance of the circuit is '$R$'. The force needed to move the rod in the direction as shown with a constant speed '$V$' is

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