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Question
A toroidal solenoid with air core has an average radius of 15 cm, area of cross-section 12 cm2 and has 1200 turns. Calculate the self-inductance of the toroid. Assume the field to be uniform across the cross-section of the toroid.
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Solution
The self inductance of a toroidal solenoid is given by \[L = \frac{\mu_0 N^2 A}{2\pi r}\]
Here, N = number of turns
A = area of cross-section
r = average radius
\[\therefore L = \frac{4\pi \times {10}^{- 7} \times \left( 1200 \right)^2 \times 12 \times {10}^{- 4}}{2\pi \times 15 \times {10}^{- 2}}\]
\[ = 230 . 4 \times {10}^{- 5} H\]
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