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184_notes:examples:week12_force_between_wires [2017/11/07 16:23] – [Solution] tallpaul | 184_notes:examples:week12_force_between_wires [2021/07/08 13:16] – schram45 | ||
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===== Magnetic Force between Two Current-Carrying Wires ===== | ===== Magnetic Force between Two Current-Carrying Wires ===== | ||
Two parallel wires have currents in opposite directions, $I_1$ and $I_2$. They are situated a distance $R$ from one another. What is the force per length $L$ of one wire on the other? | Two parallel wires have currents in opposite directions, $I_1$ and $I_2$. They are situated a distance $R$ from one another. What is the force per length $L$ of one wire on the other? | ||
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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==== | ||
- | We know that the magnetic field at the location of Wire 2 from Wire 1 is given by the magnetic field of a long wire: | + | We will start by trying to find the magnetic force on Wire 2. Since there is a current in Wire 1, we know that there will be a magnetic field in the space around that wire. There would also be a magnetic field from the current in Wire 2, but this magnetic field (from Wire 2) won't contribute to the magnetic force on Wire 2 because Wire 2 cannot exert a force on itself. This means that the force on Wire 2 is due to how the magnetic field from Wire 1 interacts with the current going through Wire 2. Mathematically, |
- | $$\vec{B}_1=\frac{\mu_0 I_1}{2 \pi r} \hat{z}$$ | + | $$ d\vec{F}_{1 \rightarrow 2} = I_2 d\vec{l}_2 \times \vec{B}_1$$ |
+ | |||
+ | We know that the magnetic field from Wire 1 at the location of Wire 2 is given by the magnetic field of a long wire: | ||
+ | $$\vec{B}_1=\frac{\mu_0 I_1}{2 \pi R} \hat{z}$$ | ||
+ | |||
+ | We can reason that the direction of the field is $+\hat{z}$ because of the [[184_notes: | ||
+ | |||
+ | Since we know the magnetic field, the next thing we wish to define is $\text{d}\vec{l}_2$. Wire 2 has current directed with $\hat{y}$ in our representation, | ||
+ | $$\text{d}\vec{l}_2 = \text{d}y \hat{y}$$ | ||
- | We can reason that the direction | + | When we take the cross product |
+ | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \text{d}y \hat{y} \times \frac{\mu_0 I_1}{2 \pi R} \hat{z}$$ | ||
+ | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \biggl(\hat{y}\times\hat{z}\biggr)$$ | ||
+ | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \hat{x}$$ | ||
- | Since we know the magnetic field, the next thing we wish to define is $\text{d}\vec{l}$. Wire 2 has current directed with $\hat{y}$ in our representation, so we can say | + | This equation gives us the force on a very small chunk of the wire. If we want to find the force on a large piece of the wire, then we have to integrate, which means we need to choose the limits for our integral. |
- | $$\text{d}\vec{l} = \text{d}y \hat{y}$$ | + | $$\vec{F}_{1 \rightarrow 2 \text{, L}} = \int_0^L I_2 \text{d}\vec{l}_2 \times \vec{B}_1$$ |
+ | $$\vec{F}_{1 \rightarrow | ||
+ | $$\vec{F}_{1 \rightarrow 2 \text{, L}} = \frac{\mu_0 I_1 I_2 L}{2 \pi R} \hat{x}$$ | ||
+ | If we wanted to write the force per length (rather than the total force), we would simply divide by L on both sides: | ||
+ | $$\frac{\vec{F}_{1 \rightarrow 2}}{L} | ||
+ | [{{ 184_notes: | ||
- | This gives | + | 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 $d\vec{l}_1$, take the cross product, integrate over a segment of the wire, divide by L). We can also use what we know about forces and Newton' |
- | $$\text{d}\vec{l} \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi r}\text{d}y \hat{x}$$ | + | |
- | 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: | + | $$\frac{\vec{F}_{2 \rightarrow 1}}{L} |
- | $$\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 | + | We can check this with the right hand rule again. First, we know that the magnetic field from Wire 2 should point out of the page at Wire 1 - point your fingers in the direction of $I_2$, curl them toward Wire 1 and then your thumb should point out of the page. We can then find the direction of the force on Wire 1 by pointing your fingers in the direction of $I_1$, curling them out of the page toward the magnetic field from Wire 2, and your thumb should point to the left (or the $-\hat{x}$ |
- | {{ }} | + | 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. |