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The segment $A B$ of wire carrying current $I_{1}$ is placed perpendicular to a long straight wire carrying current $I_{2}$ as shown in figure. The magnitude of force experienced by straight wire $A B$ is
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$\frac{μ_{0} I_{1} I_{2}}{2 π} \ln 3$

$\frac{μ_{0} I_{1} I_{2}}{2 π} \ln 2$

$\frac{2 μ_{0} I_{1} I_{2}}{2 π}$

$\frac{μ_{0} I_{1} I_{2}}{2 π}$

answer is B.

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Detailed Solution

At a distance $X$ from current $I_{2}$ ,

$B = \frac{μ_{0}}{2 π} \frac{I_{2}}{X}$

Magnetic force of small element $d X$ of wire $A B d F = I_{2} (d X) B \sin 90 °$

$∴ F = \int_{x = a}^{x = 2 a} d F$

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