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 θ 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 $B=\sqrt{\frac{mgR\mathrm{sin\theta }}{v{l}^2{\cos }^2\mathrm{\theta }}}$

Answer 1

Component of weight along its motion, F' = mgsinθ
The emf induced in the rod due to its motion is given by
e = Bl'v'
Here,
l' = Component of the length of the rod perpendicular to the magnetic field
v' = Component of the velocity of the rod perpendicular to the magnetic field
$i=\frac{B\times l\times v\mathrm{cos\theta }}{R}$
$$\begin{gathered} \left|\overrightarrow{F}\right|=i\left|\overrightarrow{l}\times \overrightarrow{B}\right|=ilB\sin (90-\theta ) \\ F=ilB=\frac{Blv\cos \theta }{R}\times l\times B\cos \theta \\ F=\frac{B^2{l}^{2}v{\cos }^2\theta }{R} \end{gathered}$$
The direction of force F is opposite to F.'
Because the rod is moving with a constant velocity, the net force on it is zero.
Thus,
F $-$ F' = 0
F = F'
or
$\frac{B^2{l}^{2}v{\cos }^2\mathrm{\theta }}{R}=mg\sin \theta$
$\therefore B=\sqrt{\frac{Rmg\sin \theta }{l^{2}v{\cos }^2\theta }}$