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--- 184_notes:examples:week12_force_between_wires [2017/11/07 16:20] – [Solution] tallpaul
+++ 184_notes:examples:week12_force_between_wires [2017/11/07 16:54] – tallpaul
@@ Line 7: / Line 7: @@
 ===Lacking===
-  * $\frac{\vec{F}}{L}$
+  * $\vec{F}_{1 \rightarrow 2 \text{, L}}$
 ===Approximations & Assumptions===
@@ Line 21: / Line 21: @@
   * We represent the situation with diagram below.
-{{ 184_notes:12_representation.png?500 |Two Wires}}
+{{ 184_notes:12_two_wires_representation.png?400 |Two Wires}}
 ====Solution====
 We know that the magnetic field at the location of Wire 2 from Wire 1 is given by the magnetic field of a long wire:
-$$\vec{B}=\frac{\mu_0 I_1}{2 \pi r} \hat{z}$$
+$$\vec{B}_1=\frac{\mu_0 I_1}{2 \pi r} \hat{z}$$
 We can reason that the direction of the field is $+\hat{z}$ because of the [[184_notes:rhr|Right Hand Rule]]. We also don't care about the magnetic field from Wire 2 at the location of Wire 2, since Wire 2 cannot exert a force on itself. Now, it remains to calculate the magnetic force.
@@ Line 33: / Line 33: @@
 This gives
-$$\text{d}\vec{l} \times \vec{B} = \frac{\mu_0 I_1}{2 \pi r}\text{d}y \hat{x}$$
+$$\text{d}\vec{l} \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi r}\text{d}y \hat{x}$$
 Lastly, we need to choose the limits on our integral. We can select our origin conveniently so that the segment of interest extends from $y=0$ to $y=L$. Now, we write:
-$$\frac{\vec{F}_{1 \rightarrow 2}}{L} = \int_0^L I_2 \text{d}\vec{l} \times \vec{B} = \int_0^L \frac{\mu_0 I_1 I_2}{2 \pi r}\text{d}y \hat{x} = \frac{\mu_0 I_1 I_2 L}{2 \pi r} \hat{x}$$
+$$\vec{F}_{1 \rightarrow 2 \text{, L}} = \int_0^L I_2 \text{d}\vec{l} \times \vec{B}_1 = \int_0^L \frac{\mu_0 I_1 I_2}{2 \pi r}\text{d}y \hat{x} = \frac{\mu_0 I_1 I_2 L}{2 \pi r} \hat{x}$$
+{{ 184_notes:12_force_per_length.png?500 |Force Per Length}}
-------------
+Notice that this force is repellent. The equal and opposite force of Wire 2 upon Wire 1 would also be repellent. Had the two current been going in the same direction, one can imagine that the two wires would attract each other.
-For now, we write $$\text{d}\vec{l} = \langle \text{d}x, -\text{d}y, 0 \rangle$$
-We write the $y$-component with a negative sign so that $\text{d}y$ can be positive. For the separation vector, we write
-$$\vec{r} = \vec{r}_{obs} - \vec{r}_{source} = \langle 0,0,0 \rangle - \langle x, y, 0 \rangle = \langle -x, -y, 0 \rangle$$
-Notice that we can rewrite $y$ as $y=-x-L$. This equation comes from the equation for a straight line, $y=mx+b$, where the slope of the line (or wire in this case) is $m=-1$ and the y-intercept of the wire is at $b=-L$. An alternate solution to this example could also be to rotate the coordinate system so that the x or y axis lines up with wire. If finding $y$ is troublesome, it may be helpful to rotate your coordinate axes.
-{{ 184_notes:9_dl_breakdown.png?600 |Breakdown of dl-vector}}
-We can use geometric arguments to say that $\text{d}y=\text{d}x$. See the diagram above for an insight into this geometric argument. 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, we can take the cross product and find the magnitude of the $\vec{r}$:
-$$\text{d}\vec{l} \times \vec{r} = \langle 0, 0, \text{d}x(L+x) - (-\text{d}x)(-x) \rangle = \langle 0, 0, L\text{d}x \rangle = L\text{d}x \hat{z}$$
-$$r^3 = (x^2 + (L+x)^2)^{3/2}$$
-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 $\text{d}x$. (Earlier we could have equally have chosen to write everything in terms of y and dy though.) We know that 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*}