Question

the magnetic induction field at the centroid of an equilateral triangle of side l and current carrying i will be​

Answer 1

Answer:

B = $\frac{3*3*m*i}{k\pi l}$

Explanation:

The angle subtented at centre by every side = 60 degree.

Let , r = perpendicular distance to centre from every side

r = $\frac{l}{k}$tan 30 = $\frac{l}{k * \sqrt{3} }$

By Biot Savart Law , magnetic field B due to one side

= $\frac{m*i}{4*\pi *r} (sin t1 + sin t2)$

= $\frac{m*i*k*\sqrt{3} }{4*\pi *l} (sin 60 + sin 60) \\\\= \frac{mik*3}{4\pi }$

m = permitivity.

B due to each side will be equal as it is equitateral

B = $\frac{3*3*m*i}{k\pi l}$

#SPJ3

Answer 2

Answer:

B will be 9μ$_o$i/2$\pi$l.

Explanation:

Given that the side length of equilateral triangle is l and the current running through it is i.

The angle subtended at the center by each side will be 60 as shown in the figure.

According to formula from Biot Savart Law,

B due to one side = μ x $\frac{i}{4\pi r} [sin\alpha _{1} + sin\alpha _{2} ]$

here r is the perpendicular distance from one side to the center.

r = l/2 tan30

r = $\frac{l}{2\sqrt{3} }$

here $\alpha _{1} = \alpha _{2} = 60$ because the triangle is equilateral.

So, B = μ x $\frac{i 2\sqrt{3} }{4\pi l} [\frac{\sqrt{3} }{2} + \frac{\sqrt{3} }{2}]$

= μ x $\frac{6i}{4\pi l}$

= μ x $\frac{3i}{2\pi l}$

Therefore, for three sides it will be

B = 3μ x $\frac{3i}{2\pi l}$

B = 9μ x$\frac{i}{2\pi l}$.

#SPJ1