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Step-by-step Solution
Step 1: Identify the Required Formula
To find the magnetic field at a point due to a finite straight current-carrying wire, we use the formula
$B = \dfrac{\mu_0}{4\pi} \dfrac{I}{b} (\cos \theta_1 + \cos \theta_2)$,
where:
$I$ is the current flowing in the wire.
$b$ is the perpendicular distance of the point from the wire.
$\theta_1$ and $\theta_2$ are the angles subtended by the two ends of the wire at the point of observation.
Step 2: Determine Geometrical Parameters of the Equilateral Triangle
Let the side of the equilateral triangle be
$a = 4.5 \times 10^{-2}\,\text{m}$.
In an equilateral triangle of side $a$, the perpendicular distance $b$ from the centroid to any side is given by
$b = \dfrac{a}{2 \tan 60^\circ} = \dfrac{a}{2 \sqrt{3}}.
Since $\tan 60^\circ = \sqrt{3}$, we get
$b = \dfrac{4.5 \times 10^{-2}}{2 \sqrt{3}}.
Step 3: Apply the Formula for One Side
For each side of the triangle, the angles at the centroid are such that
$\theta_1 = \theta_2 = 30^\circ$ (because each side subtends 60° in total about the centroid, divided equally).
Hence,
$\cos \theta_1 + \cos \theta_2 = \cos 30^\circ + \cos 30^\circ = 2 \cos 30^\circ = 2 \times \dfrac{\sqrt{3}}{2} = \sqrt{3}.
Thus, for one side, the magnetic field contribution is:
$B_{\text{one side}}
= \dfrac{\mu_0}{4\pi} \dfrac{I}{b} (\cos 30^\circ + \cos 30^\circ)
= \dfrac{\mu_0}{4\pi} \dfrac{I}{\tfrac{a}{2 \sqrt{3}}} \sqrt{3}
= \dfrac{\mu_0}{4\pi} \dfrac{2\sqrt{3}\,I}{a} \sqrt{3}.
$
Simplifying:
$B_{\text{one side}} = \dfrac{\mu_0}{4\pi} \dfrac{6\,I}{a}.
Step 4: Sum the Contributions from All Three Sides
Since the triangle has three sides, and each side’s magnetic field at the centroid points in the same (either into or out of the plane) direction because of symmetry and the right-hand rule, the total magnetic field is thrice that of one side:
$B_{\text{total}} = 3 \times B_{\text{one side}}
= 3 \times \left( \dfrac{\mu_0}{4\pi} \dfrac{6\,I}{a} \right)
= \dfrac{18\,\mu_0\,I}{4\pi\,a}.
Using $\dfrac{\mu_0}{4\pi} = 10^{-7}\,\text{H/m}$ and substituting $a=4.5\times10^{-2}\,\text{m}$ and $I=1\,\text{A}$:
$B_{\text{total}}
= 10^{-7} \times \dfrac{18 \times 1}{4.5 \times 10^{-2}}
= 10^{-7} \times \dfrac{18}{4.5 \times 10^{-2}}
= 10^{-7} \times \dfrac{18}{0.045}.
$ \dfrac{18}{0.045} = 400.
So, $B_{\text{total}} = 400 \times 10^{-7} = 4 \times 10^{-5}\,\text{T}.
Step 5: Final Answer
Therefore, the magnetic field at the center of the equilateral triangle is
$4 \times 10^{-5}\,\text{Wb/m}^2$ (or Tesla).
