Question

A coil of radius $R$ carries current $i_1$. Another concentric coil of radius $r(r<<R)$ carries current $i_2$. Places of two coils are mutually perpendicular and both the coils are free to rotate about common diameter. If the maximum kinetic energy of smaller coil when both are released is $\frac{{\mu }_0{i}_1{i}_2\pi r^{2}MR}{X({MR}^2+{mr}^2)}$. Find X?

Masses of coils are $M$ and $m$, respectively.

Answer 1

Magnetic induction at center due to current in larger coil is: $B=\frac{{\mu }_{0}i}{2R}......\left(i\right)$
Magnetic dipole moment of smaller coil: $M=\pi r^2{i}_2......\left(ii\right)$
Initially $M$ and $B$ are at ${90}^o$
So, $U_1=-MB\cos {90}^o=0$
When the coils become coplanar
$U_2=-MB\cos 0^o=-MB$
Decrease in potential energy $=U_1-U_2=MB=\frac{{\mu }_0{i}_1{i}_2\pi r^2}{2R}$
This decrease in potential energy converts into kinetic energy.

Here the kinetic energies are maximum when the coils become coplanar. The two coils become coplanar. The two coils rotate due to their mutual interaction only and if one rotates clockwise, then the other coil rotates anticlockwise.
Let ${\omega }_1$ and ${\omega }_2$ be the angular velocities of larger and smaller coils when they become coplanar.

According to law of conservation of angular momentum: $I_1{\omega }_1={\omega }_2{I}_2$
And according to law of conservation of energy: $\frac{1}{2}{I}_1{{\omega }_1}^2+\frac{1}{2}{I}_2{{\omega }_2}^2=U$
From above two equations, we get:
$\frac{1}{2}{I}_1{\left(\frac{I_2{\omega }_2}{I_1}\right)}^2+\frac{1}{2}{I}_2{{\omega }_2}^2=U\Rightarrow \frac{1}{2}{I}_2{{\omega }_2}^2(\frac{I_2}{I_1}+1)=U$
$\Rightarrow \frac{1}{2}{I}_2{{\omega }_2}^2=\frac{U{I}_1}{I_1+I_2}$

So maximum kinetic energy of smaller coil is given by:
$\Rightarrow \frac{1}{2}{I}_2{{\omega }_2}^2=\left(\frac{{\mu }_0{i}_1{i}_2\pi r^2}{2R}\right)\left(\frac{\left(1/2\right)M{R}^2}{\left(1/2\right)M{R}^2+\left(1/2\right)M{r}^2}\right)$$=\frac{{\mu }_0{i}_1{i}_2\pi r^{2}MR}{2({MR}^2+{mr}^2)}$