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Q. A square loop of side $2a$ and carrying current I is kept in $xz$ plane with its centre at origin. A long wire carrying the same current I is placed parallel to z-axis and passing through point $(0, b, 0),(b >> a)$ The magnitude of torque on the loop about $z$ -ax is will be :

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Solution:

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$F = BI 2 a =\frac{\mu_{0} I }{2 \pi r } I \times 2 a$
$F =\frac{\mu_{0} I ^{2} a }{\pi \sqrt{ b ^{2}+ a ^{2}}}$
$\tau= F \cos \theta \times 2 a$
$=\frac{\mu_{0} I^{2} a}{\pi \sqrt{b^{2}+a^{2}}} \times \frac{b}{\sqrt{b^{2}+a^{2}}} \times 2 a$
$\tau=\frac{2 \mu_{0} I^{2} a^{2} b}{\pi\left(a^{2}+b^{2}\right)}$
If $b>>$ a then $\tau=\frac{2 \mu_{0} I^{2} a^{2}}{\pi b}$
But among the given options (1) is most appropriate