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Question

(A) The induced electromotive force $(EMF)$ in the rod is $\varepsilon = B \ell v$.The induced current in the circuit is $i = \frac{\varepsilon}{R} =

Vedclass · Accepted Answer

Straight line,Parabola

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(A) The induced electromotive force $(EMF)$ in the rod is $\varepsilon = B \ell v$.The induced current in the circuit is $i = \frac{\varepsilon}{R} = \frac{B \ell v}{R}$.The magnetic retarding force acting on the rod is $f_m = i \ell B = \left(\frac{B \ell v}{R}\right) \ell B = \frac{B^2 \ell^2 v}{R}$.To move the rod with a constant velocity $v$,the external agent must apply a force $f$ equal and opposite to the magnetic retarding force:$f = f_m = \left(\frac{B^2 \ell^2}{R}\right) v$.Since $f \propto v$,the graph of force versus velocity is a straight line passing through the origin.The power supplied by the external agent is $P = f \cdot v = \left(\frac{B^2 \ell^2}{R}\right) v \cdot v = \left(\frac{B^2 \ell^2}{R}\right) v^2$.Since $P \propto v^2$,the graph of power versus velocity is a parabola.

$A$ conducting rod moves towards the right with a constant velocity $v$ in a uniform transverse magnetic field. Determine the nature of the graphs between the force applied by the external agent versus velocity and the power supplied by the external agent versus velocity.

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