I am taking an electro-magnetics course and I have these problem.

When I found the alphas values (\$\alpha_1 = \frac{3}{\sqrt{13}}\$ and \$\alpha_2 = 0\$ because of the infinity line).

The equation says that $$cos(\alpha_2) - cos(\alpha_1)$$ which make it $$1-\frac{3}{\sqrt{13}}$$

The problem is: in the solution 145m which could be found with \$1+\frac{3}{\sqrt{13}}\$, how it could be \$+\$?

This is the question

  • \$\begingroup\$ If you gave the derivation of the formula you used maybe someone could point out the error either in the example or in your derivation. Jonk in his answer seems to prove it's not in the example. No idea why it 'gained' him a -1 though... \$\endgroup\$ – HarryH Oct 4 '19 at 21:55

For each infinitesimal segment of the wire, \$\text{d}L\$, this expression holds: $$\vec{\text{d}H} =-\frac{1}{4 \pi}\frac{I\:x\:\text{d}y}{\left(x^2+y^2\right)^\frac32}\left<0,0,1\right>$$

The above follows from applying a cross-product to the distance \$r\$ for any given infinitesimal segment.

\$x_0=2\$ is the closest approach. \$y_0=3\$.

$$\begin{align*}\frac{I\: x_0}{4\pi}\int_{y=0}^{y=\infty}\frac{\text{d}y}{\left[x_0^2+\left(y-y_0\right)^2\right]^\frac32}&=\frac{I\:x_0}{4\pi}\int_{y=-3}^{y=\infty}\frac{\text{d}y}{\left(x_0^2+y^2\right)^\frac32}\\\\&=\frac{I\: x_0}{4\pi}\left[\frac{y}{x_0^2\sqrt{x_0^2+y^2}}\right]_{y=-3}^{y=\infty}\\\\&=\frac{I}{4\pi}\left[\frac{1}{x_0}-\frac{-3}{x_0\sqrt{x_0^2+\left(-3\right)^2}}\right]\\\\&=\frac{I}{4\pi}\left[\frac{1}{2}+\frac{3}{2\sqrt{13}}\right]\\\\&=\frac{1}{4\pi}\left[1+\frac{3}{\sqrt{13}}\right]\end{align*}$$

Note the minus signs cancel out in this definite integral.

Another way of saying this is that the short, additional length of wire adds to the net magnetic strength. It doesn't subtract from it.

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