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.

Answer 1

Let ABC be the equilateral triangle with side l/3 and centre M.   

\[\text{ In } ∆ AOB, \]
\[AO = \sqrt{\left( \frac{l}{3} \right)^2 - \left( \frac{I}{6} \right)^2}\]
\[ = l\sqrt{\frac{1}{9} - \frac{1}{36}} = l\sqrt{\frac{4 - 1}{36}} = l\sqrt{\frac{1}{12}}\]
\[ \therefore \text{ MO }= \frac{1}{3} \times l\sqrt{\frac{1}{12}} = \frac{l}{6\sqrt{3}}\]

The angles made by points B and C with centre M are  \[\theta_1 = 60^\circ \text{ and } \theta_2 = 60^\circ \]

Separation of the point from the wire, d = MO = \[\frac{l}{6\sqrt{3}}\]

Thus, the magnetic field due to current in wire BC is given by

\[B = \frac{\mu_0 i}{4\pi d}(\sin \theta_1 + \sin \theta_2 )\]
\[ \Rightarrow B = \frac{\mu_0 i}{4\pi\frac{l}{6\sqrt{3}}}(\sin 60 + \sin 60)\]

\[\Rightarrow B = \frac{\mu_0 i}{4\pi l}6\sqrt{3} \times \sqrt{3}\]

Now,
Net magnetic field at M = Magnetic field due to wire BC + Magnetic field due to wire CA + Magnetic field due to wire AB
Since all wires are the same  \[B_{net} = 3B\]

 

\[= \frac{27 \mu_0 i}{\pi l}\]

It is perpendicular to the plane in outward direction if the current is anticlockwise and perpendicular to the plane in inward direction if the current is clockwise.