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184_notes:examples:week12_force_between_wires [2017/11/07 16:00] – created tallpaul | 184_notes:examples:week12_force_between_wires [2017/11/07 16:54] – tallpaul | ||
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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? | ||
+ | |||
+ | ===Facts=== | ||
+ | * $I_1$ and $I_2$ exist in opposite directions. | ||
+ | * The two wires are separated by a distance $R$. | ||
+ | |||
+ | ===Lacking=== | ||
+ | * $\vec{F}_{1 \rightarrow 2 \text{, L}}$ | ||
+ | |||
+ | ===Approximations & Assumptions=== | ||
+ | * The currents are steady. | ||
+ | * The wires are infinitely long. | ||
+ | * There are no outside forces to consider. | ||
+ | |||
+ | ===Representations=== | ||
+ | * We represent the magnetic field from a very long wire as | ||
+ | $$\vec{B}=\frac{\mu_0 I}{2 \pi r} \hat{z}$$ | ||
+ | * We represent the magnetic force on a little piece of current as | ||
+ | $$\text{d}\vec{F}= I \text{d}\vec{l} \times \vec{B}$$ | ||
+ | * We represent the situation with diagram below. | ||
+ | |||
+ | {{ 184_notes: | ||
+ | |||
+ | ====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: | ||
+ | $$\vec{B}_1=\frac{\mu_0 I_1}{2 \pi r} \hat{z}$$ | ||
+ | |||
+ | 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 $\text{d}\vec{l}$. Wire 2 has current directed with $\hat{y}$ in our representation, | ||
+ | $$\text{d}\vec{l} = \text{d}y \hat{y}$$ | ||
+ | |||
+ | This gives | ||
+ | $$\text{d}\vec{l} \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi r}\text{d}y \hat{x}$$ | ||
+ | |||
+ | 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: | ||
+ | |||
+ | $$\vec{F}_{1 \rightarrow 2 \text{, L}} = \int_0^L I_2 \text{d}\vec{l} \times \vec{B}_1 = \int_0^L \frac{\mu_0 I_1 I_2}{2 \pi r}\text{d}y \hat{x} = \frac{\mu_0 I_1 I_2 L}{2 \pi r} \hat{x}$$ | ||
+ | |||
+ | {{ 184_notes: | ||
+ | |||
+ | 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. |