1.

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»


Solution :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))`


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