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A toroid is wound over a circular core. Radius of each turn is r and radius of toroid is R (>> r). The coefficient of self inductance of the toroid is given by

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$L = \frac{μ_{0} N r^{2}}{2 R}$

$L = \frac{μ_{0} N r}{2 R}$

$L = \frac{μ_{0} N r^{2}}{R}$

$L = \frac{μ_{0} N^{2} r^{2}}{2 R}$

answer is D.

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Detailed Solution

$L = \frac{ϕ}{I}, ϕ = N A B$

$B = μ_{0} n I$

$W h e r e n = \frac{N}{2 π R}$

$∴ ϕ = N π r^{2} (μ_{0} \frac{N}{2 π R} I)$

$ϕ = \frac{μ_{0} N^{2} r^{2} I}{2 R}$

$L = \frac{ϕ}{I} = \frac{μ_{0} N^{2} r^{2}}{2 R}$

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