Question

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 :

• A. $\frac{2{\mu }_0{I}^2{a}^{2}b}{\pi \left(a^2+b^2\right)}$
• B. $\frac{{\mu }_0{I}^2{a}^{2}b}{2\pi \left(a^2+b^2\right)}$
• C. $\frac{{\mu }_0{I}^2{a}^2}{2\pi b}$
• D. $\frac{2{\mu }_0{I}^2{a}^2}{\pi b}$

Answer 1

Answer: (A)

$F=BI2a=\frac{{\mu }_{0}I}{2\pi r}I\times 2a$
$F=\frac{{\mu }_0{I}^{2}a}{\pi \sqrt{b^2+a^2}}$
$\tau =F\cos \theta \times 2a$
$=\frac{{\mu }_0{I}^{2}a}{\pi \sqrt{b^2+a^2}}\times \frac{b}{\sqrt{b^2+a^2}}\times 2a$
$\tau =\frac{2{\mu }_0{I}^2{a}^{2}b}{\pi \left(a^2+b^2\right)}$
If $b>>$ a then $\tau =\frac{2{\mu }_0{I}^2{a}^2}{\pi b}$
But among the given options (1) is most appropriate