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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 [2021/07/08 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 | + | 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 | + | |
- | $$\text{d}\vec{l} \times \vec{B} = \frac{\mu_0 I_1}{2 \pi r}\text{d}y \hat{x}$$ | + | |
- | + | ||
- | 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: | + | |
- | + | ||
- | $$\frac{\vec{F}_{1 \rightarrow 2}}}{L} = \int_0^L | + | |
+ | We know that the magnetic field from Wire 1 at the location of Wire 2 is given by the magnetic field of a long wire: | ||
+ | $$\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}$$ | ||
- | For now, we write $$\text{d}\vec{l} = \langle | + | 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. |
+ | $$\vec{F}_{1 \rightarrow 2 \text{, L}} = \int_0^L I_2 \text{d}\vec{l}_2 \times \vec{B}_1$$ | ||
+ | $$\vec{F}_{1 \rightarrow 2 \text{, L}} = \int_0^L \frac{\mu_0 I_1 I_2}{2 \pi R}\text{d}y \hat{x}$$ | ||
+ | $$\vec{F}_{1 \rightarrow 2 \text{, L}} = \frac{\mu_0 I_1 I_2 L}{2 \pi R} \hat{x}$$ | ||
+ | If we wanted to write the force per length (rather than the total force), we would simply divide by L on both sides: | ||
+ | $$\frac{\vec{F}_{1 \rightarrow 2}}{L} | ||
+ | [{{ 184_notes: | ||
- | We write the $y$-component with a negative sign so that $\text{d}y$ can be positive. For the separation vector, we write | + | If instead we wanted to find the force per length on Wire 1 from Wire 2, then we could do this whole process over again (find the magnetic field from Wire 2 at the location of Wire 1, find $d\vec{l}_1$, take the cross product, integrate over a segment of the wire, divide by L). We can also use what we know about forces and Newton' |
- | $$\vec{r} = \vec{r}_{obs} - \vec{r}_{source} = \langle 0,0,0 \rangle - \langle x, y, 0 \rangle = \langle -x, -y, 0 \rangle$$ | + | |
- | Notice that we can rewrite | + | $$\frac{\vec{F}_{2 \rightarrow 1}}{L} |
- | {{ 184_notes: | + | 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 $-\hat{x}$ direction). |
- | We can use geometric arguments to say that $\text{d}y=\text{d}x$. See the diagram above for an insight into this geometric argument. We can now plug in to express $\text{d}\vec{l}$ and $\vec{r}$ in terms of $x$ and $\text{d}x$: | + | Notice |
- | $$\text{d}\vec{l} = \langle \text{d}x, | + | |
- | $$\vec{r} = \langle -x, L+x, 0 \rangle$$ | + | |
- | Now, we can take the cross product | + | |
- | $$\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 | + | |
- | \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*} | + |