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184_notes:examples:week12_force_loop_magnetic_field [2017/11/07 21:00] – tallpaul | 184_notes:examples:week12_force_loop_magnetic_field [2021/07/13 12:33] (current) – [Solution] 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 ===== | ||
- | Two parallel wires have currents in opposite directions, | + | Suppose you have a square loop (side length |
===Facts=== | ===Facts=== | ||
- | * $I_1$ and $I_2$ exist in opposite directions. | + | * The loop is a square with side-length |
- | * The two wires are separated by a distance | + | * The magnetic field is parallel to two sides of the loop, and has a magnitude |
+ | * The current in the loop is $I$. | ||
===Lacking=== | ===Lacking=== | ||
- | * $\vec{F}_{1 \rightarrow 2 \text{, L}}$ | + | * The magnetic force on the loop. |
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * The currents are steady. | + | * The current is in a steady |
- | * The wires are infinitely long. | + | * 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 magnetic field from a very long wire as | + | * We represent the magnitude of force on a current-carrying wire in a magnetic field as |
- | $$\vec{B}=\frac{\mu_0 I}{2 \pi r} \hat{z}$$ | + | $$\left| \vec{F} |
- | * We represent the magnetic | + | * We represent the situation with the diagram below. We arbitrarily choose a counterclockwise direction for the current, and convenient coordinates axes. |
- | $$\text{d}\vec{F}= | + | |
- | * We represent the situation with diagram below. | + | |
- | {{ 184_notes:12_two_wires_representation.png?400 |Two Wires}} | + | [{{ 184_notes:12_square_loop_representation.png?400 |Square Loop in a B-field}}] |
====Solution==== | ====Solution==== | ||
- | We know that the magnetic field at the location | + | 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 |
- | $$\vec{B}_1=\frac{\mu_0 I_1}{2 \pi r} \hat{z}$$ | + | |
+ | [{{ 184_notes:12_force_theta.png? | ||
+ | |||
+ | This gives the following magnitudes: | ||
+ | |||
+ | \[ | ||
+ | \left| | ||
+ | | ||
+ | | ||
+ | | ||
+ | | ||
+ | | ||
+ | \] | ||
- | 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 field, the next thing we wish to define | + | 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: |
- | $$\text{d}\vec{l} = \text{d}y \hat{y}$$ | + | |
- | This gives | + | [{{ 184_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 | + | We could also calculate |
- | $$\vec{F}_{1 \rightarrow 2 \text{, L}} = \int_0^L I_2 \text{d}\vec{l} \times \vec{B}_1 = \int_0^L \frac{\mu_0 I_1 I_2}{2 \pi r}\text{d}y \hat{x} = \frac{\mu_0 I_1 I_2 L}{2 \pi r} \hat{x}$$ | + | $$\vec{\tau} = \vec{r} \times \vec{F}$$ |
+ | Since we have two forces, we have to take the sum of the torques from those forces. | ||
+ | $$\vec{\tau}=\vec{\tau}_{left}+\vec{\tau}_{right}$$ | ||
+ | $$\vec{\tau}= \vec{r}_\text{left} \times \vec{F}_\text{left} + \vec{r}_\text{right} | ||
- | {{ 184_notes: | + | If we say the axis of rotation is in the middle of the loop, then $r_{left}$ and $r_{right}$ would both be $L/2$. We can also plug in what we just for the forces on the left and right sides of the loop. This will then give us the total torque on the loop. |
+ | $$\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} = 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 through the loop) to kinetic energy (by making the loop spin). | ||
- | 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. | + | We can check the direction of our torques by using RHR as this is also a cross product. If we put our fingers in the direction |