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 16:23] – [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 that the two wires would attract each other. | ||