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. $\frac{mg \, R \, \tan \, \theta}{B^2 l^2}$
• B. $\frac{mg \, R \, \cot\, \theta}{B^2 l^2}$
• C. $\frac{mg \, R \, \sin\, \theta}{B^2 l^2}$
• D. $\frac{mg \, R \, \cos \, \theta}{B^2 l^2}$

Answer 1

Answer: (C)

At terminal velocity, net force on rod $=m g \sin \theta$
$\Rightarrow m g \sin \theta=i B l$
Now, $I B l=\left(\frac{B v l}{R}\right) B l=\frac{B^{2} l^{2} v}{R} $
$\Rightarrow m g \sin \theta=\frac{B^{2} l^{2} v}{R} $
$\Rightarrow v=\frac{m g R \sin \theta}{B^{2} l^{2}}$