Question

A wire of length $l$ is bent in the form of an equilateral triangle and carries an electric current $i$. (a) Find the magnetic field $B$ at the centre. (b) If the wire is bent in the form of a square, what would be the value of $B$ at the centre?

Answer 1

a) $\Delta ABC$ is Equilateral

$AB=BC=CA=\ell /3$

Current $=i$

$AO=\frac{\sqrt{3}}{2}a=\frac{\sqrt{3}\times \ell }{2\times 3}=\frac{\ell }{2\sqrt{3}}$

${\varphi }_1={\varphi }_2={60}^{\circ }$

So, $MO=\frac{\ell }{6\sqrt{3}}$ as $AM:MO=2:1$

$\overrightarrow{B}$ due to $BC$ at <

$=\frac{{\mu }_{0}i}{4\pi r}(\sin {\varphi }_1+\sin {\varphi }_2)=\frac{{\mu }_{0}i}{4\pi }\times i\times 6\sqrt{3}\times \sqrt{3}=\frac{{\mu }_{0}i\times 9}{2\pi \ell }$

net $\overrightarrow{B}=\frac{9{\mu }_{0}i}{2\pi \ell }\times 3=\frac{27{\mu }_{0}i}{2\pi \ell }$

b) $\overrightarrow{B}$ due to $AD=\frac{{\mu }_{0}i\times B}{4\pi \times \ell }\sqrt{2}=\frac{8\sqrt{2}{\mu }_{0}i}{4\pi \ell }$

Net $\overrightarrow{B}=\frac{8\sqrt{2}{\mu }_{0}i}{4\pi \ell }\times 4=\frac{8\sqrt{2}{\mu }_{0}i}{\pi \ell }$