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184_notes:examples:week12_force_loop_magnetic_field [2017/11/08 02:23] – [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=== | ||
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* We represent the situation with the diagram below. We arbitrarily choose a counterclockwise direction for the current, and convenient coordinates axes. | * 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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\] | \] | ||
- | It remains to find the direction of the force, for which we will use 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 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: | ||
- | This means that the net force on the loop is $0$, the loop's center of mass won't move! However, the opposing forces on opposite sides will cause the loop to spin -- there is a torque! The calculation for the torque is shown below, with a diagram included to show visually what happens. | + | [{{ 184_notes: |
- | {{ 184_notes: | + | We could also calculate the torque on the loop, using the definition of torque $\vec{\tau} = \vec{r} \times \vec{F}$, where $\tau$ is the torque on the object, $r$ is the distance from the loop to the axis of rotation, and $F$ is the force. |
- | The calculation is here: | + | $$\vec{\tau} = \vec{r} \times \vec{F}$$ |
+ | Since we have two forces, we have to take the sum of the torques from those forces. | ||
+ | $$\vec{\tau}=\vec{\tau}_{left}+\vec{\tau}_{right}$$ | ||
+ | $$\vec{\tau}= \vec{r}_\text{left} \times \vec{F}_\text{left} + \vec{r}_\text{right} \times \vec{F}_\text{right}$$ | ||
- | $$\vec{\tau} = \vec{r} \times \vec{F} = \vec{r}_\text{left} | + | If we say the axis of rotation is in the middle of the loop, then $r_{left}$ and $r_{right}$ would both be $L/2$. We can also plug in what we just for the forces on the left and right sides of the loop. This will then give us the total torque on the loop. |
+ | $$\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} | ||
+ | This sort of rotating loop is the basis for an electrical motor. Essentially you are transferring electric energy (by providing a current through the loop) to kinetic energy (by making the loop spin). |