The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is:
[Take $$\mu_0 = 4\pi \times 10^{-7}$$ N A$$^{-2}$$]

We have to find the magnetic field produced at the centre of an equilateral triangular loop of side $$a = 1\ \text{m}$$ carrying a current $$I = 10\ \text{A}$$. The loop lies in one plane, and the required point is its centroid (the geometric centre).

For a straight, finite conductor carrying current $$I$$, the magnetic field at a point lying at a perpendicular distance $$d$$ from the conductor is given by the Biot-Savart result

$$B \;=\; \frac{\mu_0 I}{4\pi d}\,(\sin\theta_1 + \sin\theta_2),$$

where $$\theta_1$$ and $$\theta_2$$ are the angles subtended by the line segments joining the observation point to the two ends of the conductor, measured with respect to the perpendicular drawn from the point to the wire.

Because our loop is an equilateral triangle, every side contributes an identical field at the centre. We will therefore calculate the field due to one side and multiply by 3.

Step 1: Perpendicular distance $$d$$ from the centre to a side

The altitude of an equilateral triangle of side $$a$$ is

$$h \;=\; \frac{\sqrt{3}}{2}\,a.$$

The centroid divides the altitude in the ratio 2:1, so the distance from the centroid to any side is

$$d \;=\; \frac{h}{3} \;=\; \frac{\sqrt{3}}{2}\,\frac{a}{3} \;=\; \frac{\sqrt{3}}{6}\,a.$$

Substituting $$a = 1\ \text{m}$$,

$$d \;=\; \frac{\sqrt{3}}{6}\ \text{m}.$$

Step 2: Angles $$\theta_1$$ and $$\theta_2$$ for one side

Draw the perpendicular from the centre to the chosen side. Each half of the side has length $$\dfrac{a}{2} = 0.5\ \text{m}$$. Hence

$$\tan\theta_1 \;=\; \tan\theta_2 \;=\; \frac{0.5}{d} \;=\; \frac{0.5}{\sqrt{3}/6} \;=\; \frac{0.5 \times 6}{\sqrt{3}} \;=\; \frac{3}{\sqrt{3}} \;=\; \sqrt{3}.$$

Therefore $$\theta_1 = \theta_2 = 60^\circ$$, and

$$\sin\theta_1 + \sin\theta_2 \;=\; 2 \sin 60^\circ \;=\; 2 \times \frac{\sqrt{3}}{2} \;=\; \sqrt{3}.$$

Step 3: Magnetic field from one side

Substituting $$d = \dfrac{\sqrt{3}}{6}$$ and $$\sin\theta_1 + \sin\theta_2 = \sqrt{3}$$ into the Biot-Savart expression, we get

$$\begin{aligned} B_{\text{one side}} &= \frac{\mu_0 I}{4\pi d}\,(\sin\theta_1 + \sin\theta_2) \\ &= \frac{\mu_0 I}{4\pi}\; \frac{\sqrt{3}}{d} \\ &= \frac{\mu_0 I}{4\pi}\; \frac{\sqrt{3}}{\sqrt{3}/6} \\ &= \frac{\mu_0 I}{4\pi}\; 6. \\ \end{aligned}$$

Step 4: Total magnetic field at the centre

There are three identical sides, so

$$\begin{aligned} B_{\text{total}} &= 3 \times B_{\text{one side}} \\ &= 3 \times \frac{\mu_0 I}{4\pi}\; 6 \\ &= \frac{\mu_0 I}{4\pi}\; 18. \\ \end{aligned}$$

Step 5: Insert numerical values

Given $$\mu_0 = 4\pi \times 10^{-7}\ \text{N A}^{-2}$$, so

$$\frac{\mu_0}{4\pi} = 10^{-7}\ \text{T m A}^{-1}.$$

Now put $$I = 10\ \text{A}$$:

$$\begin{aligned} B_{\text{total}} &= 18 \times 10^{-7}\,I \\ &= 18 \times 10^{-7} \times 10 \\ &= 18 \times 10^{-6}\ \text{T} \\ &= 18\ \mu\text{T}. \\ \end{aligned}$$

Hence, the correct answer is Option C.