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 the \(z\)-axis and passing through the point \((0,b,0)\), (\(b\gg a\)). The magnitude of the torque on the loop about \(z\)-axis is given by :

• A. \(\frac{2\mu_0I^{2}a^{2}}{\pi b}\)
• B. \(\frac{\mu_0I^{2}a^{2}}{2\pi b}\)
• C. \(\frac{\mu_0I^{2}a^{3}}{2\pi b^{2}}\)
• D. \(\frac{2\mu_0I^{2}a^{3}}{\pi b^{2}}\)

Hint:

For \(b\gg a\) the wire field over the loop is uniform, \(B=\frac{\mu_0I}{2\pi b}\), directed along \(x\); the loop moment is \(I(2a)^{2}\) along \(y\), so \(\tau=mB\).

Answer 1

The Correct Option is A

Field of the long wire (along \(z\), at \(y=b\)) at the origin: magnitude \(B=\frac{\mu_0I}{2\pi b}\); its direction is perpendicular to both the \(z\)-axis and the line joining the wire to the origin (the \(y\)-axis), i.e. along \(\pm x\). For \(b\gg a\) it is uniform over the loop.
The loop lies in the \(XZ\) plane, so its magnetic moment is along \(y\): \(m=I(2a)^{2}=4Ia^{2}\).
\(\vec{m}\perp\vec{B}\), so the torque is \(\tau=mB=4Ia^{2}\cdot\frac{\mu_0I}{2\pi b}=\frac{2\mu_0I^{2}a^{2}}{\pi b}\), and it acts along the \(z\)-axis (\(\vec m\times\vec B\) with \(\hat y\times\hat x\)).
Answer: (1) \(\frac{2\mu_0I^{2}a^{2}}{\pi b}\)