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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

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4.6 mH

6.9 mH

2.3 mH

9.2 mH

answer is C.

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Detailed Solution

$\begin{array}{l} L = (μ_{0} n^{2}) \times (2 π R \times A) \\ = 4 π \times 10^{- 7} \times \frac{{(1200)}^{2}}{2 π \times 0 .15} \times 12 \times 10^{- 4} H \\ = \frac{2 \times 144 \times 12}{0 .15} \times 10^{- 7} H = 2 .3 m H \end{array}$

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