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184_notes:examples:week12_force_between_wires [2017/11/07 16:24] – tallpaul | 184_notes:examples:week12_force_between_wires [2017/11/07 16:54] – tallpaul | ||
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===Lacking=== | ===Lacking=== | ||
- | * $\frac{\vec{F}}{L}$ | + | * $\vec{F}_{1 \rightarrow 2 \text{, |
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
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* 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==== | ||
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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: | {{ 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. |