A square loop of side 2a carrying current I i | vedclass

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Question

(A) The magnetic field $B$ produced by the long wire at a distance $r$ is given by $B = \frac{\mu_{0} I}{2 \pi r}$.For the two sides of the loop paral

Vedclass · Accepted Answer

$\frac{2 \mu_{0} I^{2} a^{2} b}{\pi(a^{2}+b^{2})}$

Vedclass · Answer

(A) The magnetic field $B$ produced by the long wire at a distance $r$ is given by $B = \frac{\mu_{0} I}{2 \pi r}$.For the two sides of the loop parallel to the $z$-axis,the distance from the wire is $r = \sqrt{b^2 + a^2}$.The force on each of these sides is $F = B I (2a) = \frac{\mu_{0} I}{2 \pi \sqrt{b^2 + a^2}} \cdot I \cdot 2a = \frac{\mu_{0} I^2 a}{\pi \sqrt{b^2 + a^2}}$.The torque $	au$ about the $z$-axis is produced by the vertical components of these forces. The component of force perpendicular to the lever arm is $F \cos 	heta$,where $\cos 	heta = \frac{b}{\sqrt{b^2 + a^2}}$.The total torque is $	au = 2 \cdot (F \cos 	heta) \cdot a = 2 \cdot \left( \frac{\mu_{0} I^2 a}{\pi \sqrt{b^2 + a^2}} ight) \cdot \left( \frac{b}{\sqrt{b^2 + a^2}} ight) \cdot a$.Simplifying this,we get $	au = \frac{2 \mu_{0} I^2 a^2 b}{\pi (a^2 + b^2)}$.

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