Question

A conducting wire of mass $m$ slides down two smooth conducting bars, set an angle $\theta$ to the horizontal as shown in figure. The separation between the bars is $l$. The system is located in the magnetic field $B$, perpendicular to the plane of the sliding wire and bars. The constant velocity of the wire is

• A. $\frac{mgR\sin \theta }{B^2{l}^2}$
• B. $\frac{mgR\sin \theta }{B^{}{l}^3}$
• C. $\frac{mgR\theta }{B^2{l}^5}$
• D. $\frac{mgR\sin \theta }{B^{}{l}^4}$

Answer 1

The correct option is A $\frac{mgR\sin \theta }{B^2{l}^2}$

$F=Bil-mgsin\theta =\mathrm{0.....}\left(newto{n}^{\prime }ssecondlaw\right)$
$Bil=mgsin\theta$

$\Rightarrow Bl\left(\frac{Blv}{R}\right)=mgsin\theta$

$\Rightarrow v=\frac{mgRsin\theta }{B^2{l}^2}$