A current of 1 A is flowing on the sides of an equilateral triangle of side $$4.5 \times 10^{-2}$$ m. The magnetic field at the centre of the triangle will be:

The magnetic field produced by a straight conductor of finite length at a point situated at a perpendicular distance $$r$$ from the wire is given by the Biot-Savart result

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

where $$\theta_1$$ and $$\theta_2$$ are the angles made by the lines joining the point to the two ends of the conductor with the perpendicular dropped from the point to the wire.

Each side of the equilateral triangle acts as such a finite straight wire. Let the length of a side be $$a = 4.5\times10^{-2}\,\text{m}$$ and the current in every side be $$I = 1\,\text{A}$$. We first determine the perpendicular distance $$r$$ of the centre of the triangle from any side.

For an equilateral triangle, the distance of the centroid from a side is

$$r \;=\;\frac{a\sqrt{3}}{6}.$$

Substituting $$a = 4.5\times10^{-2}\,\text{m}$$, we have

$$r \;=\;\frac{4.5\times10^{-2}\,\text{m}\;\times\;\sqrt{3}}{6}.$$

Next, we find the angles $$\theta_1$$ and $$\theta_2$$. A perpendicular from the centre to any side meets that side at its midpoint, dividing the side into two segments of length $$a/2$$. In the right-angled triangle so formed,

$$\tan\theta_1 \;=\;\tan\theta_2 \;=\;\frac{a/2}{r} \;=\;\frac{a/2}{a\sqrt{3}/6} \;=\;\frac{3}{\sqrt{3}} \;=\;\sqrt{3}.$$

Hence

$$\theta_1 = \theta_2 = 60^{\circ},\qquad \sin\theta_1 = \sin\theta_2 = \sin60^{\circ} = \frac{\sqrt{3}}{2}.$$

We now apply the formula for the magnetic field due to one side:

$$\begin{aligned} B_{\text{one side}} &=\frac{\mu_0 I}{4\pi r}\,(\sin\theta_1+\sin\theta_2)\\[6pt] &=\frac{\mu_0 I}{4\pi r}\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)\\[6pt] &=\frac{\mu_0 I}{4\pi r}\,\sqrt{3}. \end{aligned}$$

The three sides form a closed equilateral loop, and by the right-hand rule each side produces a field at the centre in the same perpendicular direction. Therefore the magnitudes add directly:

$$B_{\text{total}} =3B_{\text{one side}} =3\left(\frac{\mu_0 I}{4\pi r}\,\sqrt{3}\right) =\frac{3\sqrt{3}\,\mu_0 I}{4\pi r}.$$

We now substitute $$r = a\sqrt{3}/6$$:

$$\begin{aligned} B_{\text{total}} &=\frac{3\sqrt{3}\,\mu_0 I}{4\pi}\;\frac{1}{a\sqrt{3}/6}\\[6pt] &=\frac{3\sqrt{3}\,\mu_0 I}{4\pi}\;\frac{6}{a\sqrt{3}}\\[6pt] &=\frac{18\,\mu_0 I}{4\pi a}. \end{aligned}$$

We simplify the numerical factor $$\frac{18}{4}=4.5$$ and obtain

$$B_{\text{total}} =\frac{9\,\mu_0 I}{2\pi a}.$$

Using $$\mu_0 = 4\pi\times10^{-7}\,\text{T·m/A}$$, $$I = 1\,\text{A}$$ and $$a = 4.5\times10^{-2}\,\text{m}$$, we substitute:

$$\begin{aligned} B_{\text{total}} &=\frac{9\,(4\pi\times10^{-7})\,(1)}{2\pi\,(4.5\times10^{-2})}\\[6pt] &=\frac{9\cdot4\times10^{-7}}{2\cdot4.5\times10^{-2}}\\[6pt] &=\frac{36\times10^{-7}}{9\times10^{-2}}\\[6pt] &=\frac{36}{9}\times10^{-7+2}\\[6pt] &=4\times10^{-5}\,\text{T}. \end{aligned}$$

Since $$1\,\text{tesla} = 1\,\text{Wb/m}^2$$, the magnetic field is

$$B_{\text{total}} = 4\times10^{-5}\,\text{Wb/m}^2.$$

Hence, the correct answer is Option A.