184_notes:examples:week9_current_segment [Projects & Practices in Physics]

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=====Magnetic Field from a Current Segment=====
You may have read about how to find the [[184_notes:b_current#Magnetic_Field_from_a_Very_Long_Wire|magnetic field from a very long wire of current]]. Now, what is the magnetic field from a single segment? Suppose we have the configuration shown below. Your observation point is at the origin, and the segment of current $I$ runs in a straight line from $\langle -L, 0, 0 \rangle$ to $\langle 0, -L, 0 \rangle$.

{{ 184_notes:9_current_segment_bare.png?200 |Segment of Current}}

===Facts===
  * The current in the segment is $I$.
  * The observation point is at the origin.
  * The segment stretches from from $\langle -L, 0, 0 \rangle$ to $\langle 0, -L, 0 \rangle$.

===Lacking===
  * $\vec{B}$

===Approximations & Assumptions===
  * The current is steady, and the wire segment is uniform.

===Representations===
  * We represent the Biot-Savart Law for magnetic field from a current as
$$\vec{B}= \int \frac{\mu_0}{4 \pi}\frac{I \cdot d\vec{l}\times \vec{r}}{r^3}$$
  * We represent the situation with diagram given above.

====Solution====
Below, we show a diagram with a lot of pieces of the Biot-Savart Law unpacked. We show an example $\text{d}\vec{l}$, and a separation vector $\vec{r}$. Notice that $\text{d}\vec{l}$ is directed along the segment, in the same direction as the current. The separation vector $\vec{r}$ points as always from source to observation.

{{ 184_notes:9_current_segment.png?400 |Segment of Current}}

For now, we write $$\text{d}\vec{l} = \langle \text{d}x, \text{d}y, 0 \rangle$$
and $$\vec{r} = \vec{r}_{obs} - \vec{r}_{source} = 0 - \langle x, y, 0 \rangle = \langle -x, -y, 0 \rangle$$
Notice that we can rewrite $y$ as $y=-L-x$. This is a little tricky to arrive at, but is necessary to figure out unless you rotate your coordinate axes, which would be an alternative solution to this example. If finding $y$ is troublesome, it may be helpful to rotate. We can take the derivative of both sides to find $\text{d}y=-\text{d}x$. We can now plug in to express $\text{d}\vec{l}$ and $\vec{r}$ in terms of $x$ and $\text{d}x$:
$$\text{d}\vec{l} = \langle \text{d}x, -\text{d}x, 0 \rangle$$
$$\vec{r} = \langle -x, L+x, 0 \rangle$$
Now, a couple other quantities that we see will be useful:
$$\text{d}\vec{l} \times \vec{r} = \langle 0, 0, \text{d}x(L+x) - (-\text{d}x)(-x) \rangle = \langle 0, 0, L\text{d}x \rangle = L\text{d}x \hat{z}$$
$$r^3 = (x^2 + (L+x)^2)^{3/2}$$
The last thing we need is the bounds on our integral. Our variable of integration is $x$, since we chose to express everything in terms of $x$ and $\text{d}x$. Our segment begins at $x=-L$, and ends at $x=0$, so these will be the limits on our integral. Below, we write the integral all set up, and then we evaluate using some assistance some [[https://www.wolframalpha.com/input/?i=integral+from+-L+to+0+of+L%2F(x%5E2%2B(L%2Bx)%5E2)%5E(3%2F2)+dx|Wolfram Alpha]].
\begin{align*}
\vec{B} &= \int \frac{\mu_0}{4 \pi}\frac{I \cdot d\vec{l}\times \vec{r}}{r^3} \\
        &= \int_{-L}^0 \frac{\mu_0}{4 \pi}\frac{IL\text{d}x}{(x^2 + (L+x)^2)^{3/2}}\hat{z} \\
        &= \frac{\mu_0}{2 \pi}\frac{I}{L}\hat{z}
\end{align*}