Question

A conducting rod of length L is moving in uniform magnetic field as shown in figure. Floor and wall are conducting with zero resistance. Resistance of rod is $R\Omega$ and its lower end is pulled with constant velocity v along x-axis. Rod remain in contact with wall and floor for full time with initially $\theta$ was ${90}^{\circ }$. So,

• A. when $\theta ={30}^{\circ }$, current in the rod flows from B to A.
• B. when $\theta ={30}^{\circ }$, current in the rod is $\frac{BVL}{2R}$
• C. direction of current in the rod changes when $\theta ={30}^{\circ }$
• D. potential difference across the rod measured by a voltmetre is zero.

Answer 1

Answer: (A), (B), (D)

potential difference across the rod measured by a voltmetre is zero.


The instantaneous axis of rotation is as shown in the figure.
$v=Lsin\theta \omega \Rightarrow \omega =\frac{v}{Lsin\theta }$
${\epsilon }_{OB}=\frac{B\omega (Lsin\theta {)}^2}{2}=\frac{BvLsin\theta }{2}$
${\epsilon }_{OA}=\frac{B\omega (Lcos\theta {)}^2}{2}=\frac{BvL}{2}\frac{co{s}^2\theta }{sin\theta }$
${\epsilon }_{AB}={\epsilon }_{OB}-{\epsilon }_{OA}=\frac{BvL}{2}$
Current in the circuit $=\frac{{\epsilon }_{AB}}{R}=\frac{BvL}{2R}$
Since, ${\epsilon }_{AB}>0$, current flows from A to B outside the rod, and B to A inside the rod.
$V_{AB}={\epsilon }_{AB}-iR=0$