Question

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.

Answer 1

Magnetic induction at centre due to current in
larger coil is
`B=(mu_0i_1)/(2R)`
Magnetic dipole moment of smaller coil
`M=pir^2i_2`
Initially M and B are at `90^@`
So, `U_1=-MB cos 90^@=0`
when the coils become coplanar
`U_2=-MB cos 0^@=-MB`
Decrease in potential energy =`U_1-U_2=MB=(mu_0i_1i_2pir^2)/(2R)`
This decrease in potential energy convert into kinetic energy.
Here the kinetic energies are maximum when the coils become
coplanar. The two coils rotate due to their mutual interaction
only and if one coil 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_1omega_1=I_2omega_2`
and according to law of conservation of energy
`1/2 I_1omega_1^2+1/2I_2omega_2^2=U`
From above two equations, we get
`1/2I_1((I_2omega_2)/(I_1))^2+1/2I_2omega_2^2=U implies 1/2I_1omega_2^2((I_1)/(I_2)+1))=U`
`implies 1/2I_1omega_2^2=(UI_1)/(I_1+I_2)`
So maximum kinetic energy of smaller coil is given by
`implies 1/2I_1omega_2^2=((mu_0i_1i_2pir^2)/(2R))(((1//2)MR^2)/((1//2)MR^2+(1//2)mr^2))`
`(mu_0i_1i_2pir^2MR)/(2(MR^2+mr^2))`