Question

A wire of length $l$ bent in the form of an equilateral triangle and carries an electric current $I$. Find the magnetic field due to any of the sides at the centre if the same current $I$ flows in the triangular loop.

• A. $\frac{18I}{l}\times {10}^{-7}\text{T}$
• B. $\frac{9I}{l}\times {10}^{-7}\text{T}$
• C. $\frac{3I}{l}\times {10}^{-7}\text{T}$
• D. Zero

Answer 1

The correct option is A $\frac{18I}{l}\times {10}^{-7}\text{T}$

For an equilateral triangle, the centre of the triangle will be on the $\perp$ bisector of each side.


From above geometry,

$\text{QS}=l/6$

In $△QOS$,

$\tan {30}^{\circ }=\frac{OS}{l/6}$

$OS=\frac{l}{6\sqrt{3}}$

And we know that, the magnetic field due to a finite wire at its perpendicular bisector is,

$B=\frac{{\mu }_{0}I}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2).......\left(1\right)$

From the above diagram, at point $O$,

$\sin {\theta }_1=\sin {\theta }_2=\sin {60}^{\circ }$

So, for wire segment $\text{PQ}$, magnetic field at point $O$ will be

$B=\frac{{\mu }_{0}I}{4\pi \left(OS\right)}\left(2\sin {60}^{\circ }\right)$

$B=\frac{4\pi \times {10}^{-7}I}{4\pi \left(l/6\sqrt{3}\right)}\left(\sqrt{3}\right)$

$\therefore B=\frac{18I}{l}\times {10}^{-7}\text{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.