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 \((0, b, 0)\), \((b \gt\gt a)\). The magnitude of torque on the loop about z-axis will be :

• A. \(\frac{2\mu_0 I^2 a^2}{\pi b}\)
• B. \(\frac{2\mu_0 I^2 a^2 b}{\pi(a^2 + b^2)}\)
• C. \(\frac{\mu_0 I^2 a^2 b}{2\pi(a^2 + b^2)}\)
• D. \(\frac{\mu_0 I^2 a^2}{2\pi b}\)

Hint:

Since \(b \gg a\), the wire's field over the loop is nearly uniform, \(B = \frac{\mu_0 I}{2\pi b}\), along the x-axis; torque is \(MB\) with \(M = I(2a)^2\) along y.

Answer 1

The Correct Option is A

For \(b \gg a\) the field of the long wire over the loop is almost uniform: \(B = \frac{\mu_0 I}{2\pi b}\). At the origin, this field is perpendicular to the line from the wire (along y), so it points along the x-axis.
The loop lies in the xz plane, so its magnetic moment is along the y-axis with magnitude \(M = I(2a)^2 = 4Ia^2\).
\(\vec M\) (along y) and \(\vec B\) (along x) are perpendicular, so the torque \(\vec\tau = \vec M \times \vec B\) is along z with magnitude \(\tau = MB = 4Ia^2\cdot\frac{\mu_0 I}{2\pi b} = \frac{2\mu_0 I^2 a^2}{\pi b}\).
Answer: (1) \(\frac{2\mu_0 I^2 a^2}{\pi b}\)