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184_notes:examples:week12_force_between_wires [2017/11/07 16:20] – [Solution] tallpaul | 184_notes:examples:week12_force_between_wires [2021/07/13 12:16] – 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=== | ||
- | * 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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* 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 | + | We will start by trying to find the magnetic |
- | $$\vec{B}=\frac{\mu_0 I_1}{2 \pi r} \hat{z}$$ | + | $$ d\vec{F}_{1 \rightarrow 2} = I_2 d\vec{l}_2 \times \vec{B}_1$$ |
- | + | ||
- | We 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 $\text{d}\vec{l}$. | + | |
- | $$\text{d}\vec{l} = \text{d}y \hat{y}$$ | + | |
- | + | ||
- | This gives | + | |
- | $$\text{d}\vec{l} \times \vec{B} | + | |
- | + | ||
- | Lastly, we need to choose the limits on our integral. We can select our origin conveniently so that the segment of interest extends from $y=0$ to $y=L$. Now, we write: | + | |
- | $$\frac{\vec{F}_{1 \rightarrow 2}}}{L} = \int_0^L I_2 \text{d}\vec{l} \times | + | 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, |
+ | $$\text{d}\vec{l}_2 = \text{d}y \hat{y}$$ | ||
+ | When we take the cross product of $d\vec{l}_2$ and $B_1$, this gives | ||
+ | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \text{d}y \hat{y} \times \frac{\mu_0 I_1}{2 \pi R} \hat{z}$$ | ||
+ | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \biggl(\hat{y}\times\hat{z}\biggr)$$ | ||
+ | $$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \hat{x}$$ | ||
+ | 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. Since we are looking for the force per length (rather than the total force), we can pick whatever kind of length we want - we will choose a segment of the wire with length L to find the force. For convenience, | ||
+ | $$\vec{F}_{1 \rightarrow 2 \text{, L}} = \int_0^L I_2 \text{d}\vec{l}_2 \times \vec{B}_1$$ | ||
+ | $$\vec{F}_{1 \rightarrow 2 \text{, L}} = \int_0^L \frac{\mu_0 I_1 I_2}{2 \pi R}\text{d}y \hat{x}$$ | ||
+ | $$\vec{F}_{1 \rightarrow 2 \text{, L}} = \frac{\mu_0 I_1 I_2 L}{2 \pi R} \hat{x}$$ | ||
+ | If we wanted to write the force per length (rather than the total force), we would simply divide by L on both sides: | ||
+ | $$\frac{\vec{F}_{1 \rightarrow 2}}{L} | ||
+ | [{{ 184_notes: | ||
- | For now, we write $$\text{d}\vec{l} | + | 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' |
- | We write the $y$-component with a negative sign so that $\text{d}y$ can be positive. For the separation vector, we write | + | $$\frac{\vec{F}_{2 \rightarrow 1}}{L} |
- | $$\vec{r} = \vec{r}_{obs} - \vec{r}_{source} = \langle 0,0,0 \rangle - \langle | + | |
- | Notice that we can rewrite $y$ as $y=-x-L$. This equation comes from the equation for a straight line, $y=mx+b$, where the slope of the line (or wire in this case) is $m=-1$ and the y-intercept | + | 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 |
- | {{ 184_notes: | + | Notice that this force is repellent |
- | We can use geometric arguments | + | If we apply RHR to this problem this can help us evaluate our solution. If we point our fingers |
- | $$\text{d}\vec{l} = \langle \text{d}x, -\text{d}x, 0 \rangle$$ | + | |
- | $$\vec{r} = \langle -x, L+x, 0 \rangle$$ | + | |
- | Now, we can take the cross product and find the magnitude of the $\vec{r}$: | + | |
- | $$\text{d}\vec{l} \times \vec{r} = \langle 0, 0, \text{d}x(L+x) - (-\text{d}x)(-x) \rangle = \langle 0, 0, L\text{d}x \rangle = L\text{d}x \hat{z}$$ | + | |
- | $$r^3 = (x^2 + (L+x)^2)^{3/ | + | |
- | The last thing we need is the bounds on our integral. Our variable of integration is $x$, since we chose to express everything in terms of $x$ and $\text{d}x$. (Earlier | + | |
- | \begin{align*} | + | |
- | \vec{B} &= \int \frac{\mu_0}{4 \pi}\frac{I \cdot d\vec{l}\times \vec{r}}{r^3} \\ | + | |
- | &= \int_{-L}^0 \frac{\mu_0}{4 \pi}\frac{IL\text{d}x}{(x^2 + (L+x)^2)^{3/ | + | |
- | &= \frac{\mu_0}{2 \pi}\frac{I}{L}\hat{z} | + | |
- | \end{align*} | + |