Question

A coil of radius $R$ carries current $I$. Another concentric coil of radius $r(r<<R)$ carries current $i$. Planes of two coils are mutually perpendicular and both the coil are free to rotate about common diameter. Find the maximum kinetic energy of the smaller coil when both the coils are released. Masses of coils are $M$ and $m$, respectively.

Answer 1

Work done on a magnetic dipole having moment $M$ roted by angle $\theta$ in a uniform magnetic field $B$ is

$W=MB(1-\cos \theta )$

This work is stored in the form of energy in the system and it will convert into maximum kinetic energy when it completely released.

Magnetic induction , at center due to current carrying coil is,

$B=\frac{{\mu }_0\times I}{2R}$ and

magnetic moment of smaller coil is, $M=\pi \times r^2\times i$

initially , planes of both coils are perpendicular to each other , $\theta ={90}^0$ , hence energy of the system will be,

$U=MB$

$=\frac{{\mu }_{0}Ii\pi r^2}{2R}$

When coil is released , both the coil strts to rotate about their common diameter and their KE will be maximum when tey become coplaner.

Moment of inertia of ring along its diameter is $I_i=\frac{M{R}_i^2}{2}$

since both coils rotate with their mutual interaction , angular momentum will be conserved , $\to$ $I_1{\omega }_1=I_2{\omega }_2$ ...(1)

and using conservation of energy : $U=\frac{I_1{{\omega }_1}^2}{2}+\frac{I_2{{\omega }_2}^2}{2}$

using eq 1 in this we get,

$\frac{I_2{{\omega }_2}^2}{2}=\frac{U{I}_1}{I_1+I_2}$

$=\frac{{\mu }_0\pi IiMR{r}^2}{2(M{R}^2+m{r}^2)}$