A long straight wire carries current $I_{1}$ . There is quarter ring of radius 3a present near it as shown. Find force on the quarter ring. Quarter ring is in the plane of paper and long wire is perpendicular to the plane of paper.

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$\frac{μ_{0} I_{1} I_{2}}{2 π} \ln 2$

$\frac{μ_{0} I_{1} I_{2}}{2 π} \ln$

$\frac{μ_{0} I_{1} I_{2}}{π} \ln$

answer is B.

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Detailed Solution

B due to long straight wire $= \frac{μ_{0} I_{1}}{2 π r}$
Force on the current wire $= \int_{r_{1} - a}^{r_{2} - 5 a} \frac{μ_{0} I_{1} I_{2}}{2 π r} d x$
$= \frac{μ_{0} I_{1} I_{2}}{2 π} \ln 5$

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