184_notes:examples:week9_current_segment

This is an old revision of the document!


You may have read about how to find the 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 $\langle 0, -L, 0 \rangle$.

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.

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.

picture

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: $$\text{d}\vec{l} \times \vec{r} = \langle 0, 0, \text{d}x \cdot (L+x) - (-\text{d}x)(-x) \rangle = \langle 0, 0, L\text{d}x \rangle$$

  • 184_notes/examples/week9_current_segment.1508461554.txt.gz
  • Last modified: 2017/10/20 01:05
  • by tallpaul