Question

A copper rod of mass m slides under gravity on two smooth parallel rails, with separation $l$ and set at an angle of $\theta$ with the horizontal. At the bottom, rails are joined by a resistance $R$. There is a uniform magnetic field $B$ normal to the plane of the rails, as shown in the figure. The terminal speed of the copper rod is :

• A. $\frac{mgR\tan \theta }{B^2{l}^2}$
• B. $\frac{mgR\cot \theta }{B^2{l}^2}$
• C. $\frac{mgR\sin \theta }{B^2{l}^2}$
• D. $\frac{mgR\cos \theta }{B^2{l}^2}$

Answer 1

Answer: (C)

At terminal velocity, net force on rod $=mg\sin \theta$
$\Rightarrow mg\sin \theta =iBl$
Now, $IBl=\left(\frac{Bvl}{R}\right)Bl=\frac{B^2{l}^{2}v}{R}$
$\Rightarrow mg\sin \theta =\frac{B^2{l}^{2}v}{R}$
$\Rightarrow v=\frac{mgR\sin \theta }{B^2{l}^2}$