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A torodal solenoid with an air core has ...

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.

Text Solution

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(a) `B=mu_0n_1I=(mu_0N_1I)/l=(mu_0N_1I)/(2pir)`
Total magnetic flux `phi_B=N_1BA=(mu_0N_1^2IA)/(2pir)`
but `phi_B=LI therefore L=(mu_0)N_1^2A)/(2pir)`
`L=(4pixx10^(-7)xx1200xx1200xx12xx10^(-4))/(2pixx0.15)H`
`=2.3 xx10^(-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_0N_1I)/(2r)]`
or `|e|=(mu_0N_1N_2A)/(2r)(dI)/(dt)`
or `|e|=(4pixx10^(-7)xx1200xx300xx12xx10^(-4)xx2)/(2xx0.15xx0.05)V`
=0.023 V
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