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 figure. The plane of the rails makes an angle $\theta$ with the horizontal. A vertical magnetic field B exists in the region. If the wire slides on the rails at a constant speed v, show that
$B=\sqrt{\frac{mgRsin\theta }{v{l}^{2}co{s}^2\theta }}$

Answer 1

$I=\frac{Blv}{R}=\frac{B\times l\cos \theta \times v\cos \theta }{R}=\frac{Blv}{R}{\cos }^2\theta$
$=ilB=\frac{Blv{\cos }^2\theta \times lB}{R}$
Now $F=mg\sin \theta$ (Force due to gravity which pulls downwards)
Now $\frac{B^2{l}^{2}v{\cos }^2\theta }{R}=mg\sin \theta \Rightarrow B=\sqrt{\frac{Rmg\sin \theta }{l^{2}v{\cos }^2\theta }}$