Question

A current of $1\text{A}$ is flowing through the sides of an equilateral triangle with sides $4.5\times {10}^{-2}\text{m}$. Find the magnetic field at the center.

• A. $6\times {10}^{-5}\text{T}$
• B. $2\times {10}^{-5}\text{T}$
• C. $4\times {10}^{-5}\text{T}$
• D. $\frac{4}{3}\times {10}^{-5}\text{T}$

Answer 1

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

An equilateral triangle $\text{ABC}$ is given in the figure in which $i$ current is flowing.
Given:
$i=1\text{A};a=4.5\times {10}^{-2}\text{m}$

In triangle $△OQX$,

$\tan {60}^{\circ }=\frac{QX}{OX}$

$OX=\frac{QX}{\tan {60}^{\circ }}=\frac{a/2}{\sqrt{3}}=\frac{a}{2\sqrt{3}}$

Let,
$B_1=$ magnetic field at $O$ due to $PQ$
$B_2=$ magnetic field at $O$ due to $QR$
$B_3=$ magnetic field at $O$ due to $RP$

Now calculating magnetic field due to $QR$ part at centre $O$,

$B_2=\frac{{\mu }_0}{4\pi }\frac{i}{OX}[\sin {\theta }_1+\sin {\theta }_2]$

$B_2=\frac{{\mu }_0}{4\pi }\frac{1}{\left(a/2\sqrt{3}\right)}[\sin {60}^{\circ }+\sin {60}^{\circ }]$

$B_2=\frac{{\mu }_0}{4\pi }\frac{2\sqrt{3}}{(4.5\times {10}^{-2})}\times \sqrt{3}=\frac{{10}^{-7}\times 6}{4.5\times {10}^{-2}}$

$\therefore B_2=\frac{4}{3}\times {10}^{-5}\text{T}$

Since triangle is equilateral, so all three sides create same magnetic field at the centre (direction will also be same),

$\therefore B=B_1+B_2+B_3=3B_2$

$B=3\times \frac{4}{3}\times {10}^{-5}=4\times {10}^{-5}\text{T}$

Hence, option $\left(c\right)$ is correct.