Question

A stiff metal rod of mass $\frac{1}{3}kg$ is kept over two knife edges of length $L=1m$ with a gap of $d=10cm$ between them as shown in the figure.


The rod carries a current of $16A$ (in the direction shown) and rolls along the rails without slipping due to the magnetic force produced by a uniform magnetic field of $0.5T$ perpendicular and pointing downwards. If the rod starts from rest and its speed when it leaves the rails is $\frac{k}{\sqrt{5}}m{s}^{-1}$, then the value of k (integer) is

Answer 1

Force on rod,
$F=I(\overrightarrow{d}\times \overrightarrow{B})=IdB\hat{i}$
By Work - Energy theorem,
$K.E$ (translation) $+K.E$ (rotation) = Work done
$\Rightarrow F\times L=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

$\Rightarrow IdBL=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{1}{2}m{R}^2\right){\left(\frac{v}{R}\right)}^2$

$\Rightarrow IdBL=\frac{3}{4}m{v}^2$

$\Rightarrow v=\sqrt{\frac{4IdBL}{3m}}=\sqrt{4\times 16\times 10\times {10}^{-2}\times 0.5\times 1}$
$$=\frac{4}{\sqrt{5}}m/s$$