Question

(a) A toroidal solenoid with an air core has an average radius of 15 cm, area of cross section 12 cm² and 1200 turns. Obtain the self inductance of toroid. Ignore field variation across the cross-section of the toroid.(b) A second coil of 300 turns is wound closely on the above toroid. If the current in the primary coil is increased from 0 to 2 A in 0.05s, obtain the induced emf in the second coil.

Answer 1

a) Radius r1=15x10-2m

Area A1=2x10-4m2

N1=1200

B=μ0nI= (μ0N1I)/ (2πr1)

Total flux: -N1xBxA1=N1x(μ0xN1I/2πr1) x A1

Also Total flux =LI

LI =μ0xA1x N12I/ (2πr1)

= (4πx10-7x (1200)2x 2x10-4)/ (2πx15x10-2)

=2.304x10-4 H

The self-inductance of the toroid is 2.304x10-4 H.

(b) A1=2x10-4m2

r1=15x10-2m

N1=1200

N2 =300

(dI1/ dt) = (If –Ii)/dt = (2-0)/ (0.05) =40A/s

Total flux in 2:- Φ2 =N2xB1xA1

=N2(μ0n1I1)/A1 equation (1)

Also,Φ2 =MI1 equation (2)

=> M=N2xμ0xn1xA1

e2 =M (dI1/dt) =μ0N2n1A1x40

=4πx10-7x 300x (N1/2πr1) x2x10-4x40

=0.023V

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