(a) A toroidal solenoid with an air core has an average radius of 0.15m, area of cross section `12 xx 10^(-4) m^(2)` and 1200 turns. Obtain the self inductance of the toroid. Ignore field variation across the cross section of the toroid. (b) A second coil of 300 turns is wound closely on the toroid above. If the current in the primary coil is increased from zero to 2.0 A in 0.05 s, obtain the induced emf in the secondary coil.

Explanation

(a) `B= mu_(0)n_(1)I = (mu_(0)N_(1)I)/(l)= (mu_(0)N_(1)I)/(2pi r)`
Total magnetic flux, `phi_(B)= N_(1)BA = (mu_(0)N_(1)^(2)IA)/(2pi r)`
But `phi_(B)= LI`
`therefore L= (mu_(0)N_(1)^(2)A)/(2pi r)`
`l= (4pi xx 10^(-7) xx 1200 xx 1200 xx 12 xx 10^(-4))/(2pi xx 0.15) H`
`=2.3 xx 10^(-3) H= 2.3mH`
(b) `|e| = (d)/(dt) (phi_(2)), " where " phi_(2)` is the total magnetic flux linked with the second coil.
`|e| =(d)/(dt) (N_(2)BA) = (d)/(dt) [N_(2)(mu_(0)N_(1)I)/(2pi r)A]`
or `|e| = (mu_(0)N_(1)N_(2)A)/(2pi r) (dI)/(dt)`
or `|e| = (4pi xx 10^(-7) xx 1200 xx 300 xx 12 xx 10^(-4) xx 2)/(2pi xx 0.15 xx 0.05) V`
=0.023V