Question

A toroidal solenoid with an air core has an average radius of $15 \,cm,$ area of cross-section $12\,cm^2$ and $1200$ 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 1

Answer: (C)

For a solenoid
$B=\mu_{0} n I$
where $ n=\frac{N}{2 \pi r} $
$\Rightarrow B=\frac{\mu_{0} N I}{2 \pi r}$
Flux linked with the solenoid is
$\phi=N B A $
$\Rightarrow \phi=\frac{\mu_{0} N^{2} A}{2 \pi r}$
$\Rightarrow L=\frac{\phi}{I}=\frac{\mu_{0} N^{2} A}{2 \pi r} $
$\Rightarrow L=2.3 \times 10^{-3} H$