Question

A metal rod of length $\ell$, radius $r$ and mass $m$ rests at the one end of a parallel conducting rail of length $L$, carrying constant current $I$ as shown. Now a uniform magnetic field of magnitude $B$ switched on normal to the plane of rail. The translational kinetic energy of rod when it reaches the other end without slipping is

• A. $\frac{1}{2} I \ell B L$
• B. $\frac{2}{3} I \ell B L$
• C. $\frac{3}{4} I \ell B L$
• D. $I \ell B L$

Answer 1

Answer: (B)

$\frac{2}{3} I \ell B L$

Answer 2

The magnetic force on the rod is given by $F = I \ell B$. The work done by this force over the distance $L$ is $W = F \cdot L = I \ell B L$. Since the rod rolls without slipping, its total kinetic energy is the sum of its translational and rotational kinetic energies: $KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$. For a cylinder rolling without slipping, $I = \frac{1}{2} m r^2$ and $\omega = \frac{v}{r}$. Thus, $KE = \frac{1}{2} m v^2 + \frac{1}{2} (\frac{1}{2} m r^2) (\frac{v}{r})^2 = \frac{3}{4} m v^2$. By the work-energy theorem, $W = KE$, so $I \ell B L = \frac{3}{4} m v^2$. The translational kinetic energy is $\frac{1}{2} m v^2 = \frac{2}{3} I \ell B L$.