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--- 184_notes:examples:week12_force_between_wires [2017/11/07 16:21] – [Solution] tallpaul
+++ 184_notes:examples:week12_force_between_wires [2017/11/12 17:08] – dmcpadden
@@ 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:
+We will start by trying to find the magnetic force on Wire 2. Since there is a current in Wire 1, we know that there will be a magnetic field in the space around that wire. There would also be a magnetic field from the current in Wire 2, but this magnetic field (from Wire 2) won't contribute to the magnetic force on Wire 2 because Wire 2 cannot exert a force on itself. This means that the force on Wire 2 is due to how the magnetic field from Wire 1 interacts with the current going through Wire 2. Mathematically, we would represent this as:
-$$\vec{B}_1=\frac{\mu_0 I_1}{2 \pi r} \hat{z}$$
+$$ d\vec{F}_{1 \rightarrow 2} = I_2 d\vec{l}_2 \times \vec{B}_1$$
-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.
+We know that the magnetic field from Wire 1 at the location of Wire 2 is given by the magnetic field of a long wire:
+$$\vec{B}_1=\frac{\mu_0 I_1}{2 \pi R} \hat{z}$$
-Since we know the magnetic field, the next thing we wish to define is $\text{d}\vec{l}$. Wire 2 has current directed with $\hat{y}$ in our representation, so we can say
+We can reason that the direction of the field is $+\hat{z}$ because of the [[184_notes:rhr|Right Hand Rule]]. (Point your fingers along the direction of $I_1$ and curl them towards Wire 2. Your thumb should point out of the page then.)
-$$\text{d}\vec{l} = \text{d}y \hat{y}$$
-This gives
+Since we know the magnetic field, the next thing we wish to define is $\text{d}\vec{l}_2$. Wire 2 has current directed with $\hat{y}$ in our representation, so we can say
-$$\text{d}\vec{l} \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi r}\text{d}y \hat{x}$$
+$$\text{d}\vec{l}_2 = \text{d}y \hat{y}$$
-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:
+When we take the cross product of $d\vec{l}_2$ and $B_1$, this gives
+$$\text{d}\vec{l}_2 \times \vec{B}_1 = \text{d}y \hat{y} \times \frac{\mu_0 I_1}{2 \pi R} \hat{z}$$
+$$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \biggl(\hat{x}\times\hat{z}\biggr)$$
+$$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \hat{x}$$
-$$\frac{\vec{F}_{1 \rightarrow 2}}{L} = \int_0^L I_2 \text{d}\vec{l} \times \vec{B}_1 = \int_0^L \frac{\mu_0 I_1 I_2}{2 \pi r}\text{d}y \hat{x} = \frac{\mu_0 I_1 I_2 L}{2 \pi r} \hat{x}$$
+This equation gives us the force on a very small chunk of the wire. If we want to find the force on a large piece of the wire, then we have to integrate, which means we need to choose the limits for our integral. Since we are looking for the force per length (rather than the total force), we can pick whatever kind of length we want - we will choose a segment of the wire with length L to find the force. For convenience, we will select our origin so that the segment of interest extends from $y=0$ to $y=L$. Now, we can write the total force from Wire 1 on that segment L of Wire 2:
+$$\vec{F}_{1 \rightarrow 2 \text{, L}} = \int_0^L I_2 \text{d}\vec{l}_2 \times \vec{B}_1$$
+$$\vec{F}_{1 \rightarrow 2 \text{, L}} = \int_0^L \frac{\mu_0 I_1 I_2}{2 \pi R}\text{d}y \hat{x}$$
+$$\vec{F}_{1 \rightarrow 2 \text{, L}}  = \frac{\mu_0 I_1 I_2 L}{2 \pi R} \hat{x}$$
+If we wanted to write the force per length (rather than the total force), we would simply divide by L on both sides:
+$$\frac{\vec{F}_{1 \rightarrow 2}}{L}  = \frac{\mu_0 I_1 I_2}{2 \pi R} \hat{x}$$
+{{ 184_notes:12_force_per_length.png?500 |Force Per Length}}
+If instead we wanted to find the force per length on Wire 1 from Wire 2, then we could do this whole process over again (find the magnetic field from Wire 2 at the location of Wire 1, find $d\vec{l}_1$, take the cross product, integrate over a segment of the wire, divide by L). We can also use what we know about forces and Newton's 3rd law, to say that the force from Wire 2 on Wire 1 should be equal and opposite to the force from Wire 1 on Wire 2.
-------------
+$$\frac{\vec{F}_{2 \rightarrow 1}}{L}  = - \frac{\mu_0 I_1 I_2}{2 \pi R} \hat{x}$$
+We can check this with the right hand rule again. First, we know that the magnetic field from Wire 2 should point out of the page at Wire 1 - point your fingers in the direction of $I_2$, curl them toward Wire 1 and then your thumb should point out of the page. We can then find the direction of the force on Wire 1 by pointing your fingers in the direction of $I_1$, curling them out of the page toward the magnetic field from Wire 2, and your thumb should point to the left (or the $-\hat{x}$ direction).
+Notice that this force is repellent - the force on Wire 1 would push it away from Wire 2 and the force on Wire 2 would push it away from Wire 1. 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*}