The mutual inductance between a long straight wire and a square loop of side a as shown In figure will be :

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$\frac{μ_{0} a}{2 π} ℓ n (\frac{x + a}{x})$

$\frac{μ_{0} a}{4 π} \ln (\frac{x + a}{x})$

$\frac{μ_{0} a}{π} ℓ n (\frac{x + a}{x})$

$\frac{μ_{0} a}{2 π} ℓ n (\frac{a}{x})$

answer is A.

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Detailed Solution

$\begin{array}{l} dϕ = BdA \\ = \frac{μ_{0} Iadx}{2 πx} \\ ϕ = \frac{μ_{0} Ia}{2 π} \int_{x}^{x + a} \frac{dx}{x} \\ MI = \frac{μ_{0} Ia}{2 π} \ln (\frac{x + a}{x}) \\ M = \frac{μ_{0} a}{2 π} \ln (\frac{x + a}{x}) \end{array}$

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