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 $1.33\times {10}^{-5}T$

Due to current through side AB Magnetic field at the centre O
$B_1=\frac{{\mu }_{0}I}{4\pi a}[sin{\theta }_1+sin{\theta }_2]$
As the magnetic field due to each of the three sides is the same in magnitude and direction.
$\therefore$ Total magnetic field at O is sum of all the fields.
i.e. $B=3B_1=\frac{3{\mu }_{0}I}{4\pi a}[sin{\theta }_1+sin{\theta }_2]$
Here, $tan{\theta }_1=\frac{AD}{OD}\Rightarrow tan{60}^o=\frac{\frac{l}{2}}{a}$
$\Rightarrow a=\frac{l}{2\sqrt{3}}=\frac{9\times {10}^{-2}}{2\sqrt{3}}$
Now B
$=3\times \frac{4\pi \times {10}^{-7}\times 2}{4\pi \times \frac{9\times {10}^{-2}}{2\sqrt{3}}}[sin{60}^o+sin{60}^o]$
$=\frac{4\sqrt{3}}{9}\times {10}^{-4}[\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}]$
$=1.33\times {10}^{-5}T$