Q.
Two parallel wires $A_{1}L\ and\ B_{1}\ M$ placed at a distance $w$ are connected by a resistor R and placed in a magnetic field B which is perpendicular to the plane containing the wires (see figure). Another wire $CD$ now connects the two wires perpendicularly and made to slide with velocity $v$ through distance L. The power developed is
A) $B\frac{lv}{R}$
B) $\frac{B^{2}l^{2}v^{2}}{R}$
C) $\frac{Bwv}{R}$
D) $\frac{B^{2}w^{2}v^{2}}{R}$
Explanation
(d) Induced emf $e = Blv = BWv$
Power developed $= \frac{e^{2}}{R} = \ \frac{B^{2}\ W^{2}v^{2}}{R}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.