Question

A coil of radius R carries a current `i_(1)`. Another concentric coil of radius `r (r lt lt R)` carries a current `i_(2)`. Planes of two coils are mutually perpendicular and both the coils are free to rotate about common diameter. Find maximum kinetic energy attained by the two coils, when both are released from rest. The masses of two coils are M and m respectively.
Strategy : The magnetic field due to coil of radius R is `(mu_(0)i_(1))/(2R)`. This field exerts a torque on shorter coil. An equal and opposite torque is exerted by the shorter coil on larger coil. There is no external torque acting on the system. The angular momentum remains conserved. Also the total mechanical energy remains conserved. At any instant, let `omega_(1) and omega_(2)` represent the angular velocity

Answer 1

`l_(1)omega_(1) - l_(2)omega_(2) = 0`
or `l_(1)omega_(1) = l_(2)omega_(2) = L` (say).

The kinetic energy of the coils are `k_(1) = (1)/(2)l_(1) omega_(1)^(2) = (L^(2))/(2l_(1))`
and `K_(2) = (L_(2))/(2l_(2))`
By conservation of energy `[(k_(1) + k_(2))-0] + [-M_(2)B_(1) - 0] = 0`
`-M_(2)B_(1)` is final potential energy. The kinetic energy is maximum when potential energy is minimum. Potential energy is minimum when `M_(2)` and `B_(1)` are parallel.
`rArr k_(1) + k_(2) = (i_(2) pi r^(2)) xx (mu_(0)i_(1))/(2R)`
Also `(k_(1))/(k_(2)) = (l_(2))/(l_(1))`
`rArr (k_(1))/(k_(2) + k_(1)) = (l_(2))/(l_(1) + l_(2))`
`rArr k_(1) = ((l_(2))/(l_(1) + l_(2))) ((i_(2) pi^(2)) mu_(0) i_(1))/(2R)`
`rArr k_(1) = (mu_(0))/(2R)((mr^(4) pi i_(1) i_(2))/((MR^(2) + mr^(2)))), k_(2) = (mu_(0))/(2)((MR r^(2) pi i_(1) i_(2))/(MR^(2) + mr^(2)))`