Question

A small cylindrical magnet `M` (Fig) is placed in the centre of a thin coil of radius a consisting of `N` turns. The coil in connected to a ballistic galvanometer. The active resistance of the whole circuit is equal to `R`. Find the magnetic moment of the magnet if its removel from the coil results in a charge `q` flowing through the galvanometer.

Answer 1

Let `vec(p_(m))` be the magnetic moment of the magent `M`. Then the magnetic field due to this magnet is,
`(mu_(0))/(4pi) [(3 (vec(p)_(m). vec(r)) vec(r))/(r^(5)) - (vec(p)_(m))/(r^(3))]`
The flux associated with this, when the magnet is along the axis at a distance `x` from the centre, is
`Phi = (mu_(0))/(4pi) int [(3 (vec(p)_(m). vec(r)) vec(r))/(r^(5)) - (vec(p)_(m))/(r^(3))] d vec(S) = Phi_(1) - Phi_(2)`.
where, `Phi_(2) = (mu_(0))/(4pi) p_(m) int_(0)^(a) (2pi rho d rho)/((x^(2) + rho^(2))^(3//2)) = (mu_(0)p)/(2) ((1)/(x) - (1)/(sqrt(x^(2) + a^(2))))`
and `Phi_(1) = (3 mu_(0) p_(m) x^(2))/(4pi) int_(0)^(a) (2pi rho d rho)/((x^(2) + rho^(2))^(5//2))`
`= (mu_(0) p_(m) x^(2))/(2) ((1)/(x^(3)) - (1)/((x^(2) + a^(2))^(3//2)))`
So, `Phi = (-mu_(0) p_(m) a^(2))/(2(x^(2) + a^(2))^(3//2))`

When the flux changes, an `e.m.f. -N (d Phi)/(dt)` is induced and a current `- (N)/(R) (d Phi)/(dt)` flows. The total charge `q`, flowing as the magnet is removed to infinity from `x = 0` is,
`q = (N)/(R) Phi (x = 0) = (N)/(R). (mu_(0) p_(m))/(2a)`
or, `p_(m) = (2aq R)/(N mu_(0))`