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184_notes:examples:week12_force_loop_magnetic_field [2017/11/07 22:33] – [Solution] tallpaul | 184_notes:examples:week12_force_loop_magnetic_field [2021/07/13 12:24] – schram45 | ||
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===== Force on a Loop of Current in a Magnetic Field ===== | ===== Force on a Loop of Current in a Magnetic Field ===== | ||
Suppose you have a square loop (side length $L$) of current $I$ situated in a uniform magnetic field $\vec{B}$ so that the magnetic field is parallel to two sides of the loop. What is the magnetic force on the loop of current? | Suppose you have a square loop (side length $L$) of current $I$ situated in a uniform magnetic field $\vec{B}$ so that the magnetic field is parallel to two sides of the loop. What is the magnetic force on the loop of current? | ||
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===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * The current is in a steady state. | + | * The current is in a steady state: This means the current in our loop is not changing with time or space, and is just a constant. |
- | * The magnetic field does not change. | + | * The magnetic field does not change: This removes any time or space dependency on our magnetic field. Assuming it constant in magnitude and direction across each segment of wire. Depending on what is producing this magnetic field, this could change the accuracy of this assumption. |
- | * There are no outside forces to consider. | + | * There are no outside forces to consider: We are not told anything about the mass of the wire or anything else that could produce a force on the loop. To simplify our model we will assume there are no outside forces like gravity, or other external magnetic fields acting on our loop. |
+ | * The current in the loop goes counterclockwise: | ||
===Representations=== | ===Representations=== | ||
* We represent the magnitude of force on a current-carrying wire in a magnetic field as | * We represent the magnitude of force on a current-carrying wire in a magnetic field as | ||
$$\left| \vec{F} \right|=IBL\sin\theta$$ | $$\left| \vec{F} \right|=IBL\sin\theta$$ | ||
- | * We represent the situation with the diagram below. | + | * We represent the situation with the diagram below. We arbitrarily choose a counterclockwise direction for the current, and convenient coordinates axes. |
- | {{ 184_notes: | + | [{{ 184_notes: |
====Solution==== | ====Solution==== | ||
In order to break down our approach into manageable chunks, we split up the loop into its four sides, and proceed. It is easy to find the magnitude of the force on each side, since $\theta$ for each side is just the angle between the magnetic field and the directed current. | In order to break down our approach into manageable chunks, we split up the loop into its four sides, and proceed. It is easy to find the magnitude of the force on each side, since $\theta$ for each side is just the angle between the magnetic field and the directed current. | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
This gives the following magnitudes: | This gives the following magnitudes: | ||
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\[ | \[ | ||
\left| \vec{F} \right| = \begin{cases} | \left| \vec{F} \right| = \begin{cases} | ||
- | | + | |
| | ||
| | ||
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\] | \] | ||
- | We can reason that the direction of the field is $+\hat{z}$ because of the [[184_notes: | + | It remains to find the direction of the force (the non-zero ones at least), for which we will use the [[184_notes: |
- | + | ||
- | Since we know the magnetic | + | |
- | $$\text{d}\vec{l} = \text{d}y | + | |
- | This gives | + | Since these forces point in opposite directions, this means that the net force on the loop is $0$, the loop's center of mass won't move! However, if there was an axis in the middle of the loop, the opposing forces on the opposite sides of the loop would cause the loop to spin. So there could be a [[183_notes: |
- | $$\text{d}\vec{l} \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi r}\text{d}y \hat{x}$$ | + | |
- | Lastly, we need to choose the limits on our integral. We can select our origin conveniently so that the segment of interest extends from $y=0$ to $y=L$. Now, we write: | + | [{{ 184_notes: |
- | $$\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}$$ | + | We could also calculate the torque on the loop, using the definition of torque |
- | {{ 184_notes: | + | $$\vec{\tau} = \vec{r} \times \vec{F}$$ |
+ | Since we have two forces, we have to take the sum of the torques from those forces. | ||
+ | $$\vec{\tau}=\vec{\tau}_{left}+\vec{\tau}_{right}$$ | ||
+ | $$\vec{\tau}= \vec{r}_\text{left} \times \vec{F}_\text{left} + \vec{r}_\text{right} \times \vec{F}_\text{right}$$ | ||
- | Notice that this force is repellent. The equal and opposite force of Wire 2 upon Wire 1 would also be repellent. Had the two current | + | If we say the axis of rotation |
+ | $$\vec{\tau} = \left(-\frac{L}{2} \hat{x}\right) \times \left(IBL \hat{z}\right) + \left(\frac{L}{2} \hat{x}\right) \times \left(-IBL \hat{z}\right)$$ | ||
+ | $$\vec{\tau} = IBL^2 \hat{y}$$ | ||
+ | This sort of rotating loop is the basis for an electrical motor. Essentially you are transferring electric energy (by providing a current |