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An infinite current-carrying conductor, carrying current $i$ , parallel to z–axis, is situated at point P as shown in the figure. The value of $\int_{A}^{B} \vec{B} \cdot \vec{d l}$ is given by $α \frac{μ_{0} i}{96}$ , then fill the value of $α$ in OMR sheet

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Detailed Solution

$\int_{A}^{B} \vec{B} \cdot \vec{d l} = \int_{\frac{π}{6}}^{\frac{π}{4}} (\frac{μ_{0} i}{2 π a}) (\frac{a d θ}{\cos θ}) (\cos θ)$

$= \frac{μ_{0} i}{2 π} \int_{\frac{π}{6}}^{π / 4} d θ = \frac{μ_{0} i}{2 π} . \frac{π}{12} = \frac{μ_{0} i}{24}$

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