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A toroidal solenoid with an air core has...

A toroidal 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.

Text Solution

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(a) Here, `a = 15 cm = 0.15 m`
`A = 12 cm^(2) = 12 xx 10^(-4) m^(2)`
Total no. of turns, `N_(1) = 1200`
Length of toroidal solenoid,
`l = 2 pi a = 2 pi xx 0.15 m = 0.3 pi m`
As `L = mu_(0) (N_(1)^(2))/(l) A`
`:. L = 4 pi xx 10^(-7) xx ((1200)^(2) xx 12 xx 10^(-4))/(0.3 pi)`
`= 2.304 xx 10^(-3)` henry
(b) Here, `N_(2) = 300`,
`(dI)/(dt) = (2 - 0)/(0.05) = 40 amp//sec., e = ?`
As `e = M = (dI)/(dt) = ((mu_(0) N_(1) N_(2) A))/(l) (dI)/(dt)`
`:. e = 4 pi xx 10^(-7) xx 1200 xx (300)/(0.3 pi) xx 12 xx 10^(-4) xx 40`
`= 0.023` volt
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