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--- 184_notes:examples:week9_current_segment [2017/10/20 00:51] – [Solution] tallpaul
+++ 184_notes:examples:week9_current_segment [2017/10/20 02:13] (current) – [Magnetic Field from a Current Segment] tallpaul
@@ Line 2: / Line 2: @@
 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.png?400 |Segment of Current}}
+{{ 184_notes:9_current_segment_bare.png?200 |Segment of Current}}
 ===Facts===
@@ Line 23: / Line 23: @@
 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}}
+{{ 184_notes:9_current_segment.png?400 |Segment of Current}}
 For now, we write  $\text{d}\vec{l} = \langle \text{d}x, \text{d}y, 0 \rangle$
@@ Line 30: / Line 30: @@
  $\text{d}\vec{l} = \langle \text{d}x, -\text{d}x, 0 \rangle$
  $\vec{r} = \langle -x, L+x, 0 \rangle$
+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*}