Question

A current of $1A$ is flowing along the sides of an equilateral triangle of side $4.5\times {10}^{-2}m$. The magnetic field at the centroid of the triangle is $({\mu }_0=4\pi \times {10}^{-7}H/m)$

• A. $4\times {10}^{-5}T$
• B. $2\times {10}^{-5}T$
• C. $4\times {10}^{-4}T$
• D. $2\times {10}^{-4}T$

Answer 1

The correct option is A $4\times {10}^{-5}T$

Given - $i=1A,s=4.5\times {10}^{-2},{\mu }_0=4\pi \times {10}^{-7}H/m$

The perpendicular distance of each side from centroid , will be the inner radius of equilateral triangle ,which is given by

$r=s\sqrt{3}/6$

Now , the magnetic field at any point $P$ at a distance $r$ from a current carrying straight wire of finite length is given by ,

$B=\frac{{\mu }_{0}i}{4\pi r}(\sin {\varphi }_1+\sin {\varphi }_2)$

where ${\varphi }_1$ and ${\varphi }_2$ are the angles inclined by lines joining the point $P$ and ends of wire , with the perpendicular from point $P$ to the wire .

Here , ${\varphi }_1={\varphi }_2={60}^o$ (by geometry)

Hence , the magnetic field at any point $P$ (centroid) at a distance $r$ from a current carrying straight wire (side of triangle) of finite length is given by ,

$B=\frac{4\pi \times {10}^{-7}i}{4\pi \times \left(s\sqrt{3}/6\right)}(\sin 60+\sin 60)$

or $B=\frac{{10}^{-7}\times 1\times 6}{4.5\times {10}^{-2}\sqrt{3}}(\sqrt{3}/2+\sqrt{3}/2)$

or $B=6\times {10}^{-5}/4.5T$

Therefore , total magnetic field by all three sides will be ,

$B^{\prime }=3\times 6\times {10}^{-5}/4.5=4\times {10}^{-5}T$