Question

A current of $1A$ is flowing on the sides of an equilateral triangle of side $4.5\times 10^{-2}m$. The magnetic field at the centre of the triangle will be :

• A. $2\times 10^{-5}Wb/m^2$
• B. Zero
• C. $8\times 10^{-5}Wb/m^2$
• D. $4\times 10^{-5}Wb/m^2$

Answer 1

Answer: (D)

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
$<br/>|B|=\frac{{\mu }_0}{4\pi d}\left(\sin {\theta }_1+\sin {\theta }_2\right)<br/>$
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
$<br/>|B|=\frac{3\times {\mu }_0}{4\pi d}\left(\sin {\theta }_1+\sin {\theta }_2\right)<br/>$
The distance $d$ can be expressed in terms of a by the formula $\frac{d}{a/2}=\tan 30$. Substituting for $d$ in the previous expression, we get,
$<br/>|B|=\frac{3\times {\mu }_0}{4\pi \left(\frac{1}{2\sqrt{3}}\right)}\left(\sin 60^{\circ }+\sin 60^{\circ }\right)<br/>$
$<br/>=4\times 10^{-5}T<br/>$