A coil of radius R carries current `i_1`. Another concentric coil of radius `r(rlt ltR)` carries current `i_2`. Planes of two coils are mutually perpendicular and both the coil are free to rotate about common diametre. Find maximum kinetic energy of smaller coil when both the coils are released, masses of coils are M and m, respectively.

Explanation

If a magnetic dipole having moment M be rotated through angle `theta` from equilibrium position in a uniform magnetic field B, work doen on it is `W=MG(1- cos theta)` . This work is stored in the system in the form of energy. When system is release, dipole starts to rotate to occupy equlibrium position and the energy converts into kinetic energy and kinetic energy of the system is maximum when stored is completely released.
Magnetic induction, at centres due to current in larger coil is `B=(mu_(0)I)/(2R)`
Magnetic dipole moment of smaller coil is `i pi r^(2)` .
Initially planes of two coils are mutually perpendicular , thereofre `theta` is `90^(@)` or energy of the system is
`U=(i pi r^(2))B(1-cos 90^(@)) U=(mu_(0)l i pi r^(2))/(2R)`
When coils are release , both the coils start to rotate about their common diameter and their kinetic energies are maximum when thay become coplanar . Moment of inertia of larger coil about axis of rotation is `I_(1)=(1)/(2)mR^(2)`
and that of smaller coil is `I_(2)=(1)/(2)mr^(2)`
Since, two coils rotate due to their mutual interaction only, therefore, if one coil rotates clockwise then the other rotates anticlockwise.
Let angular velocities of larger and smaller coils be numberically equal to `omega _(1)` and `omega_(2) ` respectively when they become coplanar.
According to law of conservation of angular momentum.
`I_(1)omega_(1)=I_(2)omega_(2)`
and according to law of conservation of energy.
`(1)/(2)I_(1)omega_(1)^(2)+(1)/(2)I_(2)omega_(2)^(2)=U`
From above equations, maximum kinetic energy of smaller coil,
`(1)/(2)I_(2)omega_(2)^(2)=(UI_(1))/(I_(1)+I_(2))=(mu_(0)pi li mRr^(2))/(2 (MR^(2)+mr^(2)))`