Question

A conducting wire $xy$ of length $l$ and mass $m$ is sliding without friction on vertical conduction rails $ab$ and $cd$ as shown in figure. A uniform magnetic field $B$ exists perpendicular to the plane of the rails, $x$ moves with a constant velocity of

• A. $\frac{mgR}{Bl}$
• B. $\frac{mgR}{B{l}^2}$
• C. $\frac{mgR}{B^2{l}^2}$
• D. $\frac{mgR}{B^{2}l}$

Answer 1

Answer: (C)

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

The emf of the circuit is $\epsilon =Blv$

The current, $I=\frac{\epsilon }{R}=\frac{Blv}{R}$

For equilibrium of rod, $mg=BIl=B\times \frac{Blv}{R}\times l$

$v=\frac{mgR}{B^2{l}^2}$