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| 184_notes:examples:week12_force_between_wires [2018/07/19 13:32] – curdemma | 184_notes:examples:week12_force_between_wires [2021/07/13 12:16] – schram45 | ||
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| ===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
| - | * The currents are steady. | + | * The currents are steady: There are many things that could cause the current in these wires to not be steady. Since we are not told anything except the fact these currents exist, we will assume them to be steady to simplify down the model. |
| - | * The wires are infinitely long. | + | * The wires are infinitely long: When the wires are infinitely long the magnetic field from each wire becomes only a function of radial distance away from the wire. This also lets us use a simpler equation derived in the notes for a long wire. |
| - | * There are no outside forces to consider. | + | * There are no outside forces to consider: We are not told anything about external magnetic fields or the mass of the wire, so we will omit forces due to gravity, external fields, and other forces. |
| ===Representations=== | ===Representations=== | ||
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| 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 | ||
| $$\text{d}\vec{l}_2 \times \vec{B}_1 = \text{d}y \hat{y} \times \frac{\mu_0 I_1}{2 \pi R} \hat{z}$$ | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \text{d}y \hat{y} \times \frac{\mu_0 I_1}{2 \pi R} \hat{z}$$ | ||
| - | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \biggl(\hat{x}\times\hat{z}\biggr)$$ | + | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \biggl(\hat{y}\times\hat{z}\biggr)$$ |
| $$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \hat{x}$$ | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \hat{x}$$ | ||
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| 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. | 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. | ||
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| + | If we apply RHR to this problem this can help us evaluate our solution. If we point our fingers in the direction of current and curl towards the external magnetic field we can find the resultant force direction for each wire. These directions will align with the forces calculated above and we can be confident in our solution. | ||