Question

A toroidal solenoid with an air core has an average radius of 15 cm, area if cross-section $12c{m}^2$ 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)

2.3 mH

Given, $radius=15cm,cross-section=12c{m}^2,N=1200$

The self inductance of toroid is given by:

$l=\frac{{\mu }_0{N}^{2}A}{2\pi r}=\frac{2\times {10}^{-7}(1200{)}^2\times 12\times {10}^{-4}}{0.15}=0.000023=2.3mH$