A horizontal flat coil of radius a made of w turns of wire carrying a current sets up a magnetic field. A horizontal conducting ring of radius r is placed at a distance `x_(0)` from the centre of the coil (Fig. 30.6). The ring is dropped. What e.m.f. will be established in it? Express the e.m.f. in terms of the speed.

Answer

`|epsi|=(3pimu iwa^2r^2v(x_(0)-vt))/(2|a^2+(x_(0)-vt)^2|^(5//2))`

Explanation

Since the ring is small, the field inside it may be assumed to be uniform and equal to the field on the axis. Therefore the magnetic flux is `Psi=BS=(2mu_(0)p_(m)pir^2)/(4pi (a^(2)+x^(2))^(3//2))`
The variable in thiscase is the coordinate `x=x_(0)-vt`, where is the velocity of fall. The magnitude of the individual e.m.f. is `|epsi|=(d Phi)/(dt)=(mu_(0)p_(m) r^2)/(2) (d)/(dt) (a^(2) x^2)^(-3//2)`
`=-3/4 mu_(0)p_(m)r^2 (a^2+r^2)^(-5//2). 2r (dr)/(dt)=(3mu_(0)p_(m) r^2xv)/(2(a^(2)+x^(2))^(5//2))`