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Q.

A toroidal solenoid with an air core has an average radius of 15 cm, area of cross section 12 cm2 and 1200 turns. Ignoring the field variation across the cross-section of the toroid, the self inductance of the toroid is

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a

4.6 mH

b

6.9 mH

c

2.3 mH

d

 9.2 mH

answer is C.

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Detailed Solution

L=(μ0n2)×(2πR×A)      =4π×10−7×(1200)22π×0.15×12×10−4H      =2×144×120.15×10−7H=2.3mH

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