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Question

Physics Question on Inductance

A toroidal solenoid with an air core has an average radius of 15cm,15 \,cm, area of cross-section 12cm212\,cm^2 and 12001200 turns. Ignoring the field variation across the cross-section of the toroid, the self-inductance of the toroid is

A

4.6 mH

B

6.9 mH

C

2.3 mH

D

9.2 mH

Answer

2.3 mH

Explanation

Solution

For a solenoid
B=μ0nIB=\mu_{0} n I
where n=N2πrn=\frac{N}{2 \pi r}
B=μ0NI2πr\Rightarrow B=\frac{\mu_{0} N I}{2 \pi r}
Flux linked with the solenoid is
ϕ=NBA\phi=N B A
ϕ=μ0N2A2πr\Rightarrow \phi=\frac{\mu_{0} N^{2} A}{2 \pi r}
L=ϕI=μ0N2A2πr\Rightarrow L=\frac{\phi}{I}=\frac{\mu_{0} N^{2} A}{2 \pi r}
L=2.3×103H\Rightarrow L=2.3 \times 10^{-3} H