Question

A wire of length $l$, mass $m$ and resistance $R$ slides without any friction, down the parallel conducting rails of negligible resistance (see figure). The rails are connected to each other at the bottom by a resistanceless rail parallel to the wire so that the wire and the rails form a closed rectangular conducting loop. The plane of the rails makes an angle $\theta$ with the horizontal and a uniform vertical magnetic field of induction B exists throughout the region. Find the steady state velocity of the wire.

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

Answer 1

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

For steady state velocity, net force on the wire should be zero. Therefore the force acting on the wire along the plane downwards must be balanced by some force. Let force $F^{\prime }$ balances the force down the inclined plane.
As the wire starts sliding down, flux through the loop EFGH changes due to which emf will be induced. Due to induced emf current will flow through the loop. The direction of this current will be such that it opposes the reduction in the flux. Hence the direction of current will be EFGH or anticlockwise.
Since the loop EFGH carrying current is placed in magnetic field, it will experience a force given by $F=BIl$. The direction of this force is perpendicular to the direction of both current and magnetic field as shown below.

Here $F^{\prime }=Fcos\theta$
Hence $Fcos\theta =mgsin\theta$
$\begin{array}{}\Rightarrow BIlcos\theta =mgsin\theta & \text{(1)}\end{array}$
Since motional emf, $E=BlV$
where velocity$\left(V\right)$ is in the direction perpendicular to both length and magnetic field.
Therefore $V=vcos\theta$
$\Rightarrow E=Blvcos\theta$
Current through the wire, $I=\frac{E}{R}=\frac{Blvcos\theta }{R}$
Substituting this value in $\left(1\right)$, we get
$\frac{Blvcos\theta }{R}Blcos\theta =mgsin\theta$
$v=\frac{mgsin\theta R}{B^2{l}^{2}co{s}^2\theta }$
Hence option C is correct