Question

A rod AB of mass m is placed on two smooth rails P and Q in a uniform magnetic induction B as shown in figure. The rails are connected to a constant current source of current I. Find the speed attained by rod AB when it leaves off the other end of rails.

Answer 1

$\ell \sqrt{\frac{2 I B}{m}}$

Answer 2

The magnetic force on the rod is given by $F = I \ell B$, where $\ell$ is the length of the rod (distance between the rails). This force is constant since $I$, $\ell$, and $B$ are constant. The rod starts from rest and moves a distance $\ell$ along the rails. Using the work-energy theorem, the work done by the magnetic force is converted into kinetic energy. Work done $W = F \times \ell = (I \ell B) \times \ell = I \ell^2 B$. The kinetic energy of the rod is $KE = \frac{1}{2} m v^2$. Equating work done to kinetic energy: $\frac{1}{2} m v^2 = I \ell^2 B$. Solving for $v$, we get $v = \sqrt{\frac{2 I \ell^2 B}{m}} = \ell \sqrt{\frac{2 I B}{m}}$.