Question

A wire of length $l$ is bent in the form of an equilateral triangle and carries an electric current $i$. Find the magnetic field $B$ at the centre.

• A. $\frac{3{\mu }_{0}i}{2\pi l}$
• B. $\frac{9{\mu }_{0}i}{2\pi l}$
• C. $\frac{15{\mu }_{0}i}{2\pi l}$
• D. $\frac{27{\mu }_{0}i}{2\pi l}$

Answer 1

The correct option is D $\frac{27{\mu }_{0}i}{2\pi l}$

Since this is an equilateral triangle of each side length $a=\frac{l}{3}$,
Total magnetic field at centroid $=3\times$ magnetic field (B) at centroid by any one side say BC.
$B=\frac{{\mu }_{0}I}{4\pi d}(\sin \angle COM+\sin \angle BOM)$

Both the angles are $60$ degree as per our diagram.
$d$ is perpendicular distance from the centroid to side under consideration.
$d=\frac{a\tan 30}{2}=\frac{a}{2\sqrt{3}}$

Total magnetic field, $B^{\prime }$ $=3B=3\times \frac{2\sqrt{3}{\mu }_{0}I}{4\pi a}(\sin 60+\sin 60)$
Putting $a=\frac{l}{3}$,
$B^{\prime }=3B=\frac{27{\mu }_{0}I}{2\pi l}$