A torodal solenoid with an air core has an average radius of 15 cm, area of cross-section `12 cm^(2) and 1200 turns. Obtain the self inductance of the toroid. Ignore field variations 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 e.m.f. in the second 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` , `:. 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_(2)BA) = (d)/(dt)[N_(2)(mu_(0)N_(1)I)/(2r)A]`
or `|e| = (mu_(0)N_(1)N_(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`
`= 0.023V`