(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_1I)/(l) = (mu_0 N_1 I)/(2pi r)`
Total magnetic flux ,` phi_B = N_1BA = (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 = 2.3 mH`
`(b) |e| = (d)/(dt) (phi_2) , ` where `phi_2` is the total magnetic flux linked with the second coil ,
`|e| = (d)/(dt) (N_2BA) = (d)/(dt) [ N_2 (mu_0 N_1 I)/(2r)A]`
or `|e| = (mu_0 N_1N_2 A)/(2r) (dI)/(dt)`
or `|e| = (4pi xx 10^(-7) xx 1200 xx 300 xx 12 xx 10^(-4) xx 2)/(2 xx 0.15 xx 0.05) V`
= 0.023 V