A small circular coil with a mass of m consists of N turns of fine wire and carries a current of I. The coil is located in a uniform magnetic field of B with the coil axis orginally parallel to the field direction.

a. What is the period of small angle oscilations of the coil axis around its equilibrium position, assuming no friction and no mechanical force?
b. If the coil is rotated through an angle of `theta` from its equilibrium position and then released, what will be its angular speed when it passes through equilibrium position?

Answer

(a) `2pisqrt(m/(2NIpiB))` ; (b) `sqrt((4NIpiB(1-cos theta))/m)`

Explanation

(a) Magnetic moment of coil:
`M=NIA=NIpiR^2`
Let the coil is displaced by small angle `theta`, the torque on the
coil: `tau=MB sin theta`
`implies -Ialpha=MB theta` (for small angle `sin theta~~ theta`)
`implies alpha=-((2NIpiB)/m) theta`
Comparing with `alpha=-omega^2 theta` we get `omega=sqrt((2NIpiB)/m)`
Time period:
`T=(2pi)/omega=(2pi)sqrt(m/(2NIpiB))`
(b) Apply conservation of energy: `(KE+PE)_i=(KE+PE)_f`
`implies theta 0-MB cos 10^@=1/2Iomega^2-MBcos 0^@`
`implies 1/2(1/2mR^2)omega^2=MB(1-cos theta)`
`implies 1/4mR^2omega^2=NIpiR^2B(1-cos theta)`
`implies omega=sqrt((4NIpiB(1-cos theta))/m)`.