Question

A wire ab of length $l$ , mass $m$ and resistance $R$ slides on a smooth thick pair of metallic rails joined at the bottom as shown in fig. The plane of the rails makes an angle $\theta$ with the horizontal. A vertical magnetic field $B$ exist in the region. If the wire slides on the rails at a constant speed $v$ , then the value of $B$ is -

• A. $\sqrt{\frac{mgR}{v{l}^{2}co{s}^2\theta }}$
• B. $\sqrt{\frac{mgRcos\theta }{v{l}^{2}si{n}^2\theta }}$
• C. $\sqrt{\frac{mgR}{v^2{l}^{2}si{n}^2\theta }}$
• D. $\sqrt{\frac{mgRsin\theta }{v{l}^{2}co{s}^2\theta }}$

Answer 1

Answer: (D)

The emf induced in the wire moving at a constant velocity $v$ is
$E=\left(B\cos \theta \right)lv$
So the current through the wire will be
$I=\frac{\left(B\cos \theta \right)lv}{R}$
Since the rod is moving uniformly, the net force acting on it will be zero. Balancing the forces along the direction of motion of the rod, we get
$mg\sin \theta =IlB\cos \theta$
$mg\sin \theta =\frac{vBl\cos \theta }{R}\times lB\cos \theta$
$B=\sqrt{\frac{mgR\sin \theta }{v{l}^2{\cos }^2\theta }}$