Question

A rod of mass m and length L is kept on two connection points a and b in a uniform magnetic field B. As soon as switch S is closed rod jumps to height H. Charge flown through the battery is

• A. $\frac{m}{LB}2\sqrt{2gH}$
• B. $\frac{m}{3LB}\sqrt{gH}$
• C. $\frac{m}{LB}\sqrt{2gH}$
• D. $\frac{m}{LB}\sqrt{gH}$

Answer 1

The correct option is C $\frac{m}{LB}\sqrt{2gH}$


Impulse on rod, $J=\int F.dt=m{V}_0$
$\int iLBdt=m{V}_0$
$\Rightarrow \int \frac{dq}{dt}LBdt=m{V}_0\Rightarrow \Delta qLB=m{V}_0$
$\therefore \Delta q=\frac{m{V}_0}{LB}=\frac{m}{LB}\sqrt{2gH}$