Question

A current of $2A$ is flowing in the sides of an equilateral triangle of side $9cm$ . The magnetic field at the centroid of the triangle is

• A. $1.66\times 10^{-6}T$
• B. $1.22\times 10^{-4}T$
• C. $4\times 10^{-6}T$
• D. $1.44\times 10^{-4}T$

Answer 1

Answer: (C)

The magnetic field at the centre $O$ due to the current through side $AB$ is
$B_1=\frac{{\mu }_{0}I}{4\pi a}\left[\sin {\theta }_1+\sin {\theta }_2\right]$

As the magnetic field due to each of the three sides is the same in magnitude and direction. So, the total magnetic field at $O$ is sum of all the fields.
i.e. $B=3B_1=\frac{3{\mu }_{0}I}{4\pi a}$ $\left[\sin {\theta }_1+\sin {\theta }_2\right]$
Here, $\tan {\theta }_1=\frac{AD}{OD}$
$\Rightarrow \tan 60^{\circ }=\frac{\frac{l}{2}}{a}$
$\Rightarrow a=\frac{l}{2\sqrt{3}}=\frac{9\times 10^{-2}}{2\sqrt{3}}$
Now $B=3\times \frac{4\pi \times 10^{-7}\times 2}{4\pi \times \frac{9\times 10^{-2}}{2\sqrt{3}}}$ $[\sin 60^{\circ }+\sin 60^{\circ }]$
$=3\times \frac{4\sqrt{3}}{9}\times {\left(10\right)}^{-6}\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)$
$=4\times 10^{-6}T$