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--- 184_notes:examples:week9_current_segment [2017/10/20 00:36] – [Magnetic Field from a Current Segment] tallpaul
+++ 184_notes:examples:week9_current_segment [2017/10/20 02:13] – 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?400 |Segment of Current}}
 ===Facts===
@@ Line 21: / Line 21: @@
 ====Solution====
-We begin by cracking open the Biot-Savart Law. In order to find magnetic field, we will need to take a cross product $\vec{v}\times \vec{r}$. Notice that $\vec{r}$ indicates a separation vector, directed from source (point particle) to observation (location 1, 2, or 3). Since our source is at the origin of our axes, then the separation vectors are actually just $\vec{r}_{sep} = \vec{r}_{obs} - 0 = \vec{r}_{obs}$, which are just the r-vectors listed in our "Facts". Below, we have calculated the cross product for each of our three locations:
+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.
-\begin{align*}
-\vec{v}\times \vec{r}_1 &= 0 \\
-\vec{v}\times \vec{r}_2 &= 1 \text{ m}^2\text{s}^{-1} \hat{z} \\
-\vec{v}\times \vec{r}_3 &= 1 \text{ m}^2\text{s}^{-1} \hat{z}
-\end{align*}
-Note that the first cross-product is 0 because location 1 is situated perfectly in line with the straight-line path of the particle. There is no magnetic field at location 1 at all! The other two cross-products are the same because locations 2 and 3 are equidistant from this path. Though they are difference distances from the particle itself, the $\hat{x}$ portion of location 3's separation vector does not contribute at all to the cross-product.
-Next, we find the magnitudes of $r^3$, since that is another quantity we need to know in the Biot-Savart Law.
+{{ 184_notes:9_current_segment.png?400 |Segment of Current}}
-\begin{align*}
-{r_2}^3 &= 0.125 \text{ m}^3 \\
-{r_3}^3 &= 0.354 \text{ m}^3
-\end{align*}
-We don't including location 1 above since we already know the magnetic field is 0 at that location! Below, we give the magnetic field at all three locations.
+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}_1 &= \frac{\mu_0}{4 \pi}\frac{q\vec{v}\times \vec{r}_1}{{r_1}^3} = 0 \\
+\vec{B} &= \int \frac{\mu_0}{4 \pi}\frac{I \cdot d\vec{l}\times \vec{r}}{r^3} \\
-\vec{B}_2 &= \frac{\mu_0}{4 \pi}\frac{q\vec{v}\times \vec{r}_2}{{r_2}^3} = 12 \text{ nT } \hat{z} \\
+        &= \int_{-L}^0 \frac{\mu_0}{4 \pi}\frac{IL\text{d}x}{(x^2 + (L+x)^2)^{3/2}}\hat{z} \\
-\vec{B}_3 &= \frac{\mu_0}{4 \pi}\frac{q\vec{v}\times \vec{r}_3}{{r_3}^3} = 4.2 \text{ nT } \hat{z}
+        &= \frac{\mu_0}{2 \pi}\frac{I}{L}\hat{z}
 \end{align*}