Question

A current $1A$ is flowing in the sides of equilateral triangle of side $4.5\times {10}^{-2}m$ .The magnetic field at centroid of the triangle is

• A. $4\times {10}^{-5}T$
• B. $40T$
• C. $0.4\times {10}^{-5}T$
• D. $4\times {10}^{-2}T$

Answer 1

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

The triangle shown here can be considered to be having three finite wires of length a each. We need to find the magnetic field at a point on the perpendicular bisector of this wire.

The magnetic field due to a finite wire at a distance d from the wire, making angles ${\theta }_1$ and ${\theta }_2$ w.r.t the ends of the wire is given by

$\left|B\right|=\frac{{\mu }_0}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$

This point is the same for all the three wires and at a distance of x from the wire. The direction of the magnetic field at that point is inwards due to all the wires. Thus, if B is the magnetic field due to one wire, then total magnetic field due to three wires will be 3B acting inwards

$\left|B\right|=\frac{3\times {\mu }_0}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$
The distance d can be expressed in terms of a by the formula $\frac{d}{a/2}=tan30$ . Substituting for d in the previous expression, we get,

$\left|B\right|$=$\frac{3\times {\mu }_0}{4\pi \left(\frac{1}{2\sqrt{3}}\right)}(\sin {60}^o+\sin {60}^o)$
= $4\times {10}^{-5}T$