Question

A stiff metal rod is kept over two knife edges of length $L=1m$ 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 rod starts from rest and its speed when it leaves the rails is $\frac{k}{\sqrt{5}}m{s}^{-1}$, then the value of $k$ is _______

Answer 1

Answer: 4

Force on rod, $F=I(d\times B)=IdB(\hat{i})$
By work-energy theorem
$KE($ Translation $)+KE($ Rotation $)=$ Work done
$\Rightarrow F\times L=\frac{1}{2}m{v}^2+\frac{1}{2}\left(I{\omega }^2\right)$
$\Rightarrow IdBL=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{1}{2}m{R}^2\right){\left(\frac{v}{R}\right)}^2$
$\Rightarrow IdBL=\frac{3}{4}m{v}^2$
Or $v=\left(\frac{4IdBL}{3m}\right)$
$=\sqrt{\frac{4\times 16\times \left(10\times 10^{-2}\right)\times 0.5\times 1}{\left(3\times \frac{1}{3}\right)}}$
$=\sqrt{64\times \frac{1}{2}\times 10^{-1}}=4\times \frac{1}{\sqrt{5}}m{s}^{-1}$