B-Field on Axis of Equilateral Triangle

[peroAlex]

Homework Statement

I came across a pretty interesting question that asks for magnetic flux density (B-field) on the axis of the equilateral triangle. This axis is meant to be perpendicular to triangle's surface passing through its centroid. Assuming that a triangle has sides denoted $a$ and conducts current $I$.

Homework Equations

Law of Biot-Savart: $dB = \frac{\mu_0 I}{4 \pi} \frac{dl \times R}{R^3}$.
B-field of a line segment: $B = \frac{\mu_0 I}{4 \pi \rho} (\cos(\alpha_1) - \cos{\alpha_2})$

The Attempt at a Solution

I managed to compute B-field at triangle's centroid. Since $h = \frac{a \sqrt{3}}{2}$ and distance from a side to centroid is $\frac{h}{3}$, I managed to derive (for one side) $$ B_1 = \frac{\mu_0 I}{4 \pi \frac{\frac{a \sqrt{3}}{2}}{3}} (\cos{30} - \cos{150}) = ... = \frac{3 \mu_0 I}{2 \pi a}$$ so for all three sides combined the result is simply three times higher value: $B_T = 3 B_1 = \frac{9 \mu_0 I}{2 \pi a}$.

But this is where I'm stuck. I don't know how to derive for axis of triangle, but I did manage to pull out correct result which should be $\frac{ 3 \mu_0 I a^2 }{ 8 \pi \sqrt{3z^2 + a^2} (z^2 + \frac{a^2}{12}) }$. Can somebody please help me with this one or at least point me on the correct path?

By the way, I hope you're having a fantastic Tuesday!

 

[Charles Link]

It's basically a geometry problem. The $B$ field from each of the sides are equal in amplitude, but each one points at an angle $\theta$ to the z-axis, so you need to compute the component of this in the z-direction and multiply by 3. The components perpendicular to the z axis cancel by symmetry. $\\$ Meanwhile the distance $\rho=R$ in the denominator can easily be computed as well as the $(sin(\theta_1)-sin(\theta_2)$ term. (You used $cos(\alpha_1)-cos(\alpha_2)$ for this above). $\\$ Anyway, I concur with the answer that you posted=I got the algebra to work.