Question

A current of $2A$ is flowing in the sides of an equilateral triangle of side $9\,cm$ . 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, $\tan \theta _{1} = \frac{A D}{O D}$
$\Rightarrow \, \, \tan60^\circ =\frac{\frac{l}{2}}{a} \, $
$\Rightarrow \, \, \, a=\frac{l}{2 \sqrt{3}}=\frac{9 \times 10^{- 2}}{2 \sqrt{3}}$
Now $B=3\times \frac{4 \pi \times 10^{- 7} \times 2}{4 \pi \times \frac{9 \times 10^{- 2}}{2 \sqrt{3}}}$ $\left[\right.\sin 60^\circ +\sin 60 ^\circ \left]\right. \, $
$=3\times \frac{4 \sqrt{3}}{9}\times \left(10\right)^{- 6}\left(\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\right)$
$=4\times 10^{- 6}T$