Question

A copper rod of mass $m$ slides under gravity on two smooth parallel rails, with separation $1$ 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. $\frac{mgR\cos \theta }{B^2{l}^2}$
• B. $\frac{mgR\sin \theta }{B^2{l}^2}$
• C. $\frac{mgR\tan \theta }{B^2{l}^2}$
• D. $\frac{mgR\cot \theta }{B^2{l}^2}$

Answer 1

Answer: (B)

$\frac{mgR\sin \theta }{B^2{l}^2}$

$\in =\frac{d\varphi }{dt}=\frac{d\left(BA\right)}{lt}$

$=\frac{d\left(Bl\right)}{dt}$

$=\frac{Bdl}{dt}=BVl$

$F=ilB=\left(\frac{BV}{R}\right)\left(l^{2}B\right)=\frac{B^2{l}^{2}V}{R}$

At equilibrium

$\Rightarrow mg\sin \theta =\frac{B^{2}lV}{R}$

$\Rightarrow V=\frac{mgR\sin \theta }{B^2{l}^2}$