A rectangular coil (Dimension 5 cm $$\times$$ 2.5 cm) with 100 turns, carrying a current of 3 A in the clock-wise direction, is kept centered at the origin and in the X-Z plane. A magnetic field of 1 T is applied along X-axis. If the coil is tilted through 45° about Z-axis, then the torque on the coil is:

We are given a rectangular coil having dimensions $$5\ \text{cm}\times 2.5\ \text{cm}$$, number of turns $$N=100$$ and current $$I=3\ \text{A}$$ (clock-wise). The coil is first kept in the X-Z plane with its centre at the origin and a uniform magnetic field $$\mathbf B$$ of magnitude $$1\ \text T$$ is applied along the +X-axis. The coil is then rotated through $$45^{\circ}$$ about the Z-axis. We have to find the torque acting on the coil in this new position.

First we calculate the area of one turn of the coil. Converting centimetres to metres, we write

$$5\ \text{cm}=5\times 10^{-2}\ \text m=0.05\ \text m,$$

$$2.5\ \text{cm}=2.5\times 10^{-2}\ \text m=0.025\ \text m.$$

Hence the area of the rectangle is

$$A = l\times b = 0.05\ \text m \times 0.025\ \text m = 0.00125\ \text{m}^2.$$

Now we discuss the angle between the coil’s area vector and the magnetic field.

• When the coil lies in the X-Z plane, its plane is perpendicular to the Y-axis, so its area vector $$\mathbf{\hat n}$$ is initially along the ±Y-direction.
• The uniform magnetic field $$\mathbf B$$ is along the X-axis.
Thus, before rotation, $$\mathbf{\hat n}$$ is at $$90^{\circ}$$ to $$\mathbf B$$.

The coil is now rotated through $$45^{\circ}$$ about the Z-axis. Rotating about the Z-axis keeps Z unchanged but swings the X and Y directions into each other. Therefore the area vector, which was along the Y-axis, now makes an angle $$45^{\circ}$$ with the Y-axis and also $$45^{\circ}$$ with the X-axis inside the X-Y plane. Consequently, the angle $$\theta$$ between the area vector and the magnetic field (which is along X) becomes

$$\theta = 45^{\circ}.$$

The magnitude of the torque acting on a current-carrying coil in a uniform magnetic field is given by the well-known formula

$$\tau = N I A B \sin\theta,$$

where

$$N = 100,\; I = 3\ \text A,\; A = 0.00125\ \text{m}^2,\; B = 1\ \text T,\; \theta = 45^{\circ}.$$

Substituting these values we have

$$\tau = 100 \times 3\ \text A \times 0.00125\ \text{m}^2 \times 1\ \text T \times \sin 45^{\circ}.$$

We know that $$\sin 45^{\circ} = \dfrac{1}{\sqrt 2} \approx 0.7071.$$ Therefore,

$$\tau = 100 \times 3 \times 0.00125 \times 1 \times 0.7071.$$

First multiply the current with the area:

$$3 \times 0.00125 = 0.00375.$$

Next include the number of turns:

$$100 \times 0.00375 = 0.375.$$

Finally include the factor $$\sin 45^{\circ}$$:

$$0.375 \times 0.7071 \approx 0.265.$$

Thus the magnitude of the torque is

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

Among the given choices, the value $$0.27\ \text{N m}$$ corresponds to Option D.

Hence, the correct answer is Option D.