Question

A toroidal solenoid with an air core has an average radius of $15 \,cm$, area if crosssection $12\, cm ^{2}$ 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)

2.3 mH

Answer 2

Given, radius $=15 \,cm$, cross $-$ section $=12 \,cm ^{2},$ $N =1200$ The self inductance of toroid is given by: $1=\frac{\mu_{0} N ^{2} A }{2 \pi r }=\frac{2 \times 10^{-7}(1200)^{2} \times 12 \times 10^{-4}}{0.15}$ $=0.000023=2.3 \,mH$