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 [2018/07/19 13:31] – curdemma | 184_notes:examples:week12_force_between_wires [2021/07/08 13:16] – schram45 | ||
---|---|---|---|
Line 23: | Line 23: | ||
* We represent the situation with diagram below. | * We represent the situation with diagram below. | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
====Solution==== | ====Solution==== | ||
Line 39: | Line 39: | ||
When we take the cross product of $d\vec{l}_2$ and $B_1$, this gives | When we take the cross product of $d\vec{l}_2$ and $B_1$, this gives | ||
$$\text{d}\vec{l}_2 \times \vec{B}_1 = \text{d}y \hat{y} \times \frac{\mu_0 I_1}{2 \pi R} \hat{z}$$ | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \text{d}y \hat{y} \times \frac{\mu_0 I_1}{2 \pi R} \hat{z}$$ | ||
- | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \biggl(\hat{x}\times\hat{z}\biggr)$$ | + | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \biggl(\hat{y}\times\hat{z}\biggr)$$ |
$$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \hat{x}$$ | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \hat{x}$$ | ||
Line 48: | Line 48: | ||
If we wanted to write the force per length (rather than the total force), we would simply divide by L on both sides: | 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{\vec{F}_{1 \rightarrow 2}}{L} | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
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 $d\vec{l}_1$, | 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 $d\vec{l}_1$, |