Question

In the figure magnetic field points into the plane of paper and the conducting rod of length $l$ is moving in this field such that the lowest point has a velocity $v_1$ and the topmost point has the velocity $v_2(v_2>v_1)$. The emf induced is given by :

• A. $B{v}_{1}l$
• B. $B{v}_{2}l$
• C. $\frac{1}{2}B(v_2+v_1)l$
• D. $\frac{1}{2}B(v_2-v_1)l$

Answer 1

The correct option is C $\frac{1}{2}B(v_2+v_1)l$

Velocity of a point at a distance $x$ from the bottom is given by:

$v=v_1+\left(\frac{v_2-v_1}{l}\right)x$
Potential difference on a small length at this distance is $e=Bv\left(dx\right)$
$\therefore$ Total potential difference $e={\int }_0^{l}Bvdx$

$e={\int }_0^{l}B(v_1+\frac{(v_2-v_1)x}{l})dx$

$=\frac{B(v_1+v_2)l}{2}$