Question

A current of $1A$ is flowing on the sides of an equilateral triangle of side $4.5\times {10}^{-2}m$. The magnetic field at the centre of the triangle will be:

• A. $4\times {10}^{-5}Wb/m^2$
• B. Zero
• C. $2\times {10}^{-5}Wb/m^2$
• D. $8\times {10}^{-5}Wb/m^2$

Answer 1

Answer: (A)

$4\times {10}^{-5}Wb/m^2$

$l=4.5\times {10}^{-2}m$

$\tan {60}^o=\sqrt{3}=\frac{l}{2d}$

$\Rightarrow d=\frac{l}{2\sqrt{3}}=\left(\frac{4.5\times {10}^{-2}}{2\sqrt{3}}\right)m$

$B=\frac{{\mu }_{o}i}{4\pi d}(\cos {\theta }_1+\cos {\theta }_2)$

$=\frac{2{\mu }_{0}i}{4\pi d}\left(\frac{\sqrt{3}}{2}\right)$

$=\frac{{\mu }_{o}i}{2\pi }\left(\frac{\sqrt{3}}{2}\right)\frac{2\sqrt{3}}{(4.5\times {10}^{-2})}$

On solving we will get option A as answer