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Q.
A current of $1\, A$ 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 :
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
$
| B |=\frac{\mu_{0}}{4 \pi d }\left(\sin \theta_{1}+\sin \theta_{2}\right)
$
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 $3 B$ acting inwards
$
| B |=\frac{3 \times \mu_{0}}{4 \pi d }\left(\sin \theta_{1}+\sin \theta_{2}\right)
$
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,
$
| B |=\frac{3 \times \mu_{0}}{4 \pi\left(\frac{1}{2 \sqrt{3}}\right)}\left(\sin 60^{\circ}+\sin 60^{\circ}\right)
$
$
=4 \times 10^{-5} T
$