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 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:-
Updated On: Jan 30, 2025
- 4.6 mH
- 6.9 mH
- 2.3 mH
- 9.2 mH
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The Correct Option is C
Solution and Explanation
Given, radius $=15 \,cm$,
cross $-$ section $=12 \,cm ^{2}, $
$N =1200$
The self inductance of toroid is given by:
$1=\frac{\mu_{0} N ^{2} A }{2 \pi r }=\frac{2 \times 10^{-7}(1200)^{2} \times 12 \times 10^{-4}}{0.15}$
$=0.000023=2.3 \,mH$
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