Question

A square loop of side L carries a current I. Another smaller square loop of side l [ l < < L ) carrying a current i is placed inside the bigger loop such that they are coplanar with their centres coinciding. If the currents in the loops are in the same direction the magnitude of the torque on the smaller loop is

• A. $\frac{\mu_0 Iil^2}{\sqrt2 \pi L}$
• B. $\frac{\mu_0 Iil^2}{2 \pi L}$
• C. $\frac{\mu_0 Iil^2}{\sqrt3 \pi L}$
• D. Zero

Answer 1

Answer: (D)

Zero

Answer 2

Given:
Larger square loop with side length L and current I.
Smaller square loop with side length l (where $$$l\lt\lt L$ and current i, placed inside the larger loop such that their centers coincide and currents are in the same direction.

To find: The magnitude of the torque $\tau$ on the smaller loop.
Magnetic Field at the Center of the Smaller Loop:
The magnetic field B at the center of the smaller square loop, due to the larger square loop carrying current I, is approximately:
$B = \frac{\mu_0 I}{2} \left( \frac{1}{\sqrt{2}} + \ln\left(1 + \sqrt{2}\right) \right)$

Magnetic Moment of the Smaller Loop:
The magnetic moment $\mu$ of the smaller square loop with current i is:
$\mu = i \cdot l^2$

Torque Calculation:
The torque $\tau$ experienced by the smaller loop is given by:
$\tau = \mu \cdot B \cdot \sin \theta$
Since the currents in both loops are in the same direction, the magnetic moment $\mu$ and the magnetic field B are parallel, so $\theta = 0^\circ$ and $\sin \theta = 0$.

Therefore,
$\tau = \mu \cdot B \cdot \sin \theta = i \cdot l^2 \cdot B \cdot 0 = 0$

So, the correct option is (D): Zero