Question

A current of $2\,A$ 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^{-5} T $
• B. $1.22 \times 10^{-4} T $
• C. $1.33 \times 10^{-5} T $
• D. $1.44 \times 10^{-4} T $

Answer 1

The Correct Option is C

The magnetic field at the centre O due to the current through side AB is
$B_1 = \frac{\mu_0 I}{4 \pi a} [\sin \, \theta_1 + \sin \, \theta_2]$

As the magnetic field due to each of the three sides is the same in magnitude and direction. So, the total magnetic field atO 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 \tan60^{\circ} = \frac{\frac{l}{2}}{a} $
$ \Rightarrow a = \frac{l}{2\sqrt{3}} = \frac{9\times10^{-2}}{2\sqrt{3}} $
Now B $ = 3 \times\frac{4\pi\times10^{-7} \times2 }{4 \pi\times\frac{9 \times10^{-2}}{2\sqrt{3}}} \, \, \, \, [\sin \, 60^{\circ} + \sin 60^{\circ} ]$
$ = \frac{4\sqrt{3}}{9} \times10^{-5} \left[ \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\right] $
$=\frac{4}{3} \times 10^{-5}$
$ = 1.33 \times10^{-5} T $