1. Magnetic torque on a current loop

For a coil of $N$ turns carrying current $I$, the magnetic moment is

$\mu = NIA$

and the torque in a magnetic field $\vec B$ is

$\tau = \mu B \sin\theta$

where $\theta$ is the angle between $\vec\mu$ and $\vec B$.

2. Area of the rectangular coil

Dimensions are

$5\text{ cm} = 0.05\text{ m}, \qquad 2.5\text{ cm} = 0.025\text{ m}$

So area,

$A = 0.05 \times 0.025 = 1.25 \times 10^{-3}\text{ m}^2$

3. Magnetic moment of the coil

Given:

• $N = 100$
• $I = 3\text{ A}$
• $A = 1.25\times10^{-3}\text{ m}^2$

Thus,

$\mu = NIA = 100 \times 3 \times 1.25\times10^{-3} = 0.375\text{ A m}^2$

4. Initial orientation of the coil

The coil lies in the $X$-$Z$ plane, so its area vector (magnetic moment direction) is along the $Y$-axis.

Since current is clockwise when viewed appropriately, $\vec\mu$ is along either $+Y$ or $-Y$; for torque magnitude only the direction sign does not matter.

Magnetic field is along the $X$-axis.

Initially, angle between $\vec\mu$ and $\vec B$ is $90^\circ$.

5. After tilting through $45^\circ$ about the $Z$-axis

Rotation about the $Z$-axis turns the magnetic moment from the $Y$-axis toward the $X$-axis by $45^\circ$.

Hence the angle between $\vec\mu$ and $\vec B$ becomes

$\theta = 45^\circ$

Therefore,

$\tau = \mu B \sin 45^\circ$

with $B=1\text{ T}$.

So,

$\tau = 0.375 \times 1 \times \frac{1}{\sqrt 2}$

$\tau \approx 0.375 \times 0.707$

$\tau \approx 0.265\text{ N m}$

$\tau \approx 0.27\text{ N m}$

6. Option check

• A: $0.42\text{ N m}$ ❌
• B: $0.55\text{ N m}$ ❌
• C: $0.38\text{ N m}$ ❌
• D: $0.27\text{ N m}$ ✅

Final Answer: $\boxed{0.27\text{ N m}}$ (Option D)