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(D) When the conductor falls with velocity $v$,an induced $emf$ $E = Blv$ is generated across its ends.Since the conductor is connected to a resistanc

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

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(D) When the conductor falls with velocity $v$,an induced $emf$ $E = Blv$ is generated across its ends.Since the conductor is connected to a resistance $R$,an induced current $I = E/R = Blv/R$ flows through the circuit.Due to this current,a magnetic force $F_m = IlB$ acts on the conductor in the upward direction.Substituting the value of $I$,we get $F_m = (Blv/R) \cdot lB = \frac{B^2l^2v}{R}$.The conductor is under the influence of two forces: the downward gravitational force $mg$ and the upward magnetic force $F_m$.The terminal speed is attained when the net force on the conductor becomes zero,i.e.,when the upward magnetic force equals the downward gravitational force:$mg = F_m$$mg = \frac{B^2l^2v}{R}$Solving for $v$,we get the terminal speed:$v = \frac{mgR}{B^2l^2}$

Similar Questions

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