Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision Next revisionBoth sides next revision | ||
184_notes:examples:week12_force_between_wires [2017/11/07 16:21] – [Solution] tallpaul | 184_notes:examples:week12_force_between_wires [2017/11/07 16:54] – tallpaul | ||
---|---|---|---|
Line 7: | Line 7: | ||
===Lacking=== | ===Lacking=== | ||
- | * $\frac{\vec{F}}{L}$ | + | * $\vec{F}_{1 \rightarrow 2 \text{, |
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
Line 21: | Line 21: | ||
* We represent the situation with diagram below. | * 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==== | ====Solution==== | ||
Line 37: | Line 37: | ||
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: | 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: | ||
- | $$\frac{\vec{F}_{1 \rightarrow 2}}{L} = \int_0^L I_2 \text{d}\vec{l} \times \vec{B}_1 = \int_0^L \frac{\mu_0 I_1 I_2}{2 \pi r}\text{d}y \hat{x} = \frac{\mu_0 I_1 I_2 L}{2 \pi r} \hat{x}$$ | + | $$\vec{F}_{1 \rightarrow 2 \text{, L}} = \int_0^L I_2 \text{d}\vec{l} \times \vec{B}_1 = \int_0^L \frac{\mu_0 I_1 I_2}{2 \pi r}\text{d}y \hat{x} = \frac{\mu_0 I_1 I_2 L}{2 \pi r} \hat{x}$$ |
+ | {{ 184_notes: | ||
- | ------------ | + | 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 |
- | + | ||
- | + | ||
- | + | ||
- | 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 $m=-1$ | + | |
- | + | ||
- | {{ 184_notes: | + | |
- | + | ||
- | 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/ | + | |
- | 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:// | + | |
- | \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/ | + | |
- | &= \frac{\mu_0}{2 \pi}\frac{I}{L}\hat{z} | + | |
- | \end{align*} | + |