Question

A wire of length $l$, mass $m$ and resistance $R$ slide without any friction down the parallel conducting rails of negligible resistance. The rails are connected to each other at the bottom by a rail of negligible resistance which is parallel to the wire . The wire and rail forms a closed rectangular conducting loop. The plane of the rails makes an angle $\theta$ with the horizontal and a uniform magnetic field $B$ which exists throughout the region. Find the steady velocity of the wire.

• A. $\frac{mg\text{sin}\theta }{R{B}^2{l}^2{\cos }^2\theta }$
• B. $\frac{mg{\sin }^2\theta }{R{B}^2{l}^2{\cos }^2\theta }$
• C. $\frac{mgR\text{sin}\theta }{B^2{l}^2{\cos }^2\theta }$
• D. $\frac{mgR{\sin }^2\theta }{B^2{l}^2{\cos }^2\theta }$

Answer 1

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


Let the steady velocity of the rod be $v$.
If we split the velocity into two componets, only $v\cos \theta$ will be responsible for induced emf in the rod as the the other component is anti parallel to it.
$ϵ=Bv\cos \theta l$
According to Ohms's law, $iR=Bv\cos \theta l$
$i=\frac{Bv\cos \theta l}{R}$
Due to this induced current, rod will experience a force in the upward direction along the incline plane.
Force due to induced current, $F_m=ilb\cos \theta$
Now to move with steady velocity
$ilB\cos \theta =mg\sin \theta$
$\frac{Bv\cos \theta l}{R}lB\theta =mg\sin \theta$
$v=\frac{mgR\text{sin}\theta }{B^2{l}^2{\cos }^2\theta }$
For detailed solution watch the next video.