Question

A wire of length l is bent in the form of an equilateral triangle and carries an electric current i. (a) Find the magnetic field B at the centre. (b) If the wire is bent in the form of a square, what would be the value of B at the centre?

Answer 1

Explanation:

Let ABC be a triangle equilateral with side $\frac{l}{3}$ and center M.

(a)  In the case of AOB,

$A O=\sqrt{\left(\frac{l}{3}\right)^{2}-\left(\frac{l}{6}\right)^{2}}$

$A O=\sqrt{\frac{l^{2}}{9}-\frac{l^{2}}{36}}$

$A O=\sqrt{l^{2}\left(\frac{1}{9}-\frac{1}{36}\right)}$

$A O=l \sqrt{\left(\frac{1}{9}-\frac{1}{36}\right)}$

$A O=l \sqrt{\left(\frac{4-1}{36}\right)}$

$A O=l |\left(\frac{3}{36}\right)$

$A O=l |\left(\frac{1}{12}\right)$

$\therefore M O=\frac{1}{3} \times l \sqrt{\left(\frac{1}{12}\right)}$

          $=\frac{l}{6 \sqrt{3}}$

The angles of points B and C with center M are the

$\theta_{1}=60^{\circ} \text { and } \theta_{2}=60^{\circ}$

The point is isolated from the wire, $\mathrm{d}=M O=\frac{i}{6 \sqrt{3}}$

Therefore, the current magnetic field in wire BC is given by

$B=\frac{\mu_{0} i}{4 \pi d}\left(\sin \theta_{1}+\sin \theta_{2}\right)$

$B=\frac{\mu_{0} i}{4 \pi \frac{l}{6 \sqrt{3}}}(\sin 60+\sin 60)$

$B=\frac{\mu_{0} i}{4 \pi l} 6 \sqrt{3} \times \sqrt{3}$

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 similar,

$B_{n e t}=3 B=\frac{27 \mu_{0} i}{\pi l}$

If the current is anticlockwise and it is perpendicular to the plane in inward direction, if the current is in clockwise direction, it is perpendicular to the plane in outward direction.  

(b)   The angles of points B and C with center M are the

$\theta_{1}=45^{\circ} \text { and } \theta_{2}=45^{\circ}$  

the point is isolated from the wire, $\mathrm{d}=\frac{1}{8}$

Therefore, the current magnetic field in wire BC is given by

$\beta=\frac{\mu_{0} t}{4 \pi d}\left(\sin \theta_{1}+\sin \theta_{2}\right)$

$B=\frac{\mu_{0} i}{4 \pi \frac{l}{8}}(\sin 45+\sin 45)$

   $=\frac{2 \sqrt{2} \mu_{0} i}{\pi l}$

Because all wires are the same, M = 4  

Net magnetic field attributable to BC wire $B_{\text {net}}=4 B=\frac{8 \sqrt{2} \mu_{0} i}{\pi l}$

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}$

Check the above attachment