Question

Consider a pair of smooth metallic rails joined at one of the ends. Rails are parallel and are inclined at ${30}^o$ with horizontal. A jumper of mass m, length l and resistance R slides down the rails with constant speed v. Magnetic field in the region is vertical. Strength of magnetic field is?

• A. $\sqrt{\frac{mgR}{3v{l}^2}}$
• B. $\sqrt{\frac{2mgR}{3v{l}^2}}$
• C. $\sqrt{\frac{3mgR}{v{l}^2}}$
• D. $\sqrt{\frac{\sqrt{3}mgR}{v{l}^2}}$

Answer 1

The correct option is B $\sqrt{\frac{2mgR}{3v{l}^2}}$

$mg\sin \theta =ilB\cos {30}^o$
$\Rightarrow \frac{mg}{2}=\frac{B^{2}lv{\cos }^2{30}^o}{R}$
$\sqrt{\frac{2mgR}{3l^v}}=B$.