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Mutual inductance in Fig. shown is -

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Zero

$\frac{μ_{0} b}{2 π} \log_{e} \frac{a}{b}$

$\frac{μ_{0} b}{2 π} \log_{e} (1 + \frac{b}{a})$

$\frac{μ_{0} b}{2 π} \log_{e} (1 + \frac{a}{b})$

answer is C.

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Detailed Solution

$B = \frac{μ_{0} I}{2 π y} a n d d ϕ = B b d y = \frac{μ_{0} I}{2 πy} . b . d y \Rightarrow ϕ = \frac{μ_{0} I b}{2 π} \int_{a}^{a + b} y^{- 1} d y = \frac{μ_{0} I b}{2 π} \ln (1 + \frac{b}{a})$

Therefore M = $\frac{ϕ}{I}$

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