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184_notes:induced_current [2021/04/08 16:00] – dmcpadden | 184_notes:induced_current [2021/11/12 23:15] (current) – [Step 1.) Draw a picture of your situation] stumptyl | ||
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We will use the example of a bar magnet moving away from a wire coil to highlight these steps and to show how you can use a table to keep track of your work. While the specifics of the table will change depending on the context, the structure and steps will work no matter what problem you are solving. So to get started you should make a table like the one shown to the right with 8 columns. The first column will be for a picture/ | We will use the example of a bar magnet moving away from a wire coil to highlight these steps and to show how you can use a table to keep track of your work. While the specifics of the table will change depending on the context, the structure and steps will work no matter what problem you are solving. So to get started you should make a table like the one shown to the right with 8 columns. The first column will be for a picture/ | ||
- | [{{184_notes: | + | [{{184_notes: |
+ | This video will walk you through an example of how to use this table or you can read about it in the notes below. | ||
+ | {{youtube> | ||
+ | \\ | ||
+ | **NOTE: In this chart there is a mistake in the chart being used: The variables " | ||
==== Right Hand Rule Steps ==== | ==== Right Hand Rule Steps ==== | ||
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Before anything, you should start with a picture of your situation. We'll draw this in the first column of the table. For our example, we'll have a bar magnet that is moving away from a set of coils. In the picture, we have marked the coils, the orientation of the magnet (which side is north/ | Before anything, you should start with a picture of your situation. We'll draw this in the first column of the table. For our example, we'll have a bar magnet that is moving away from a set of coils. In the picture, we have marked the coils, the orientation of the magnet (which side is north/ | ||
- | [{{ 184_notes:ic_scenario.png?440| Step 1: First make a diagram of the particular situation include the coil, magnet, and relevant directions.}}] | + | [{{184_notes: |
+ | }}] | ||
====Step 2.) Draw the direction of the B-field through the relevant area==== | ====Step 2.) Draw the direction of the B-field through the relevant area==== | ||
- | Next we need to determine the direction of the magnetic field through the relevant area. For this situation, the relevant area is going to be our coils so we are particularly interested in the direction of the B-field through the coils. Remember for a bar magnet, the magnetic field should point out from the north side of the magnet, wrap around, and point into the south side of the magnet. | + | Next we need to determine the direction of the magnetic field through the relevant area. For this situation, the relevant area is going to be our coils, so we are particularly interested in the direction of the B-field through the coils. Remember for a bar magnet, the magnetic field should point out from the north side of the magnet, wrap around, and point into the south side of the magnet. |
+ | [{{184_notes: | ||
- | When given a magnetic bar, remember how the B-field flows. The B-field moves from the south pole to the north pole, then wraps around the magnetic bar as it flows back to the south. This B-field is also mirrored across the bar and will show a symmetrical flow to the B-field through a magnet. | ||
- | [{{ 184_notes: | ||
- | \\ | ||
- | [{{ 184_notes: | ||
- | \\ | ||
- | ====Step | + | ====Step |
- | \\ | + | Remember that the d→A is perpendicular to the cross section area of the coils. Meaning, that you can think of the d→A as pointing |
- | Remember that the d→A is perpendicular to the cross section area of the solenoid. Meaning, that you can think of the d→A as pointing | + | |
- | \\ | + | [{{184_notes: |
- | [{{ 184_notes:ic_flux.png | + | |
- | ?440|Step 3 of the solution process. This step takes the movement of the object and generalizes the change in flux of the scenario. These are then placed into the chart.}}] | + | |
- | \\ | + | |
+ | ====Step 4.) ΦB,i, | ||
+ | Now, we're ready to figure out what is happening to the initial flux, final flux, and therefore what is happening to the change in flux for the scenario. For this step, we don't need to do exact calculations. Instead, what we really care about is the relative magnitude of the flux (is it a big or small flux) and the sign of the flux (is it positive or negative? | ||
- | ====Step 3.) dΦBdt==== | + | The magnitude |
- | \\ | + | ΦB=∫→B∙d→A=∫B∗dA∗cos(θ) |
- | This portion | + | So if θ |
- | When referring to positive/ | + | ===Initial flux=== |
+ | The initial flux is determined for the " | ||
- | \\ | + | ===Final Flux=== |
+ | The final flux is determined after some time has passed. Think about this as looking at the “End of the Scenario”. For our situation, this means the the bar magnet will have moved further away from the coil to the left. This means that magnetic field through the coil will still be pointing to the left (toward the south end of the magnet) but it will be smaller in magnitude since the magnet is further away. (Nothing has changed about our coil so the dA vector remains the same). This means that our final magnetic flux will be relatively small (compared to our initial flux), but the flux should still be positive since B and dA still point in the same direction. Thus in the fifth column, we write that the final flux is "small and positive." | ||
- | ====Initial flux==== | + | ===Change in Flux=== |
- | Now that we have an understanding of the key terms we can start to walk through | + | Now we can determine |
- | \\ | + | $$\Delta \Phi_B = \Phi_{B,f} - \Phi_{B,i}$$ |
- | ====Final Flux==== | + | |
- | The final flux is calculated __AFTER THE POSITION CHANGE__ of the object. Think about this as looking at the “End of the Scenario”. Compared to the initial position, if the objects are FARTHER, then we would deem this final flux to be “small”. If the objects are now CLOSER, then we would deem the final flux to be “large”. | + | |
+ | In our case, this means we'd be taking a small positive number minus a big positive number. This will result in a // | ||
+ | [{{184_notes: | ||
+ | ?440|Step 4: Determine the sign of the change in flux based on the initial and final flux for the situation. | ||
- | [{{ 184_notes:ic_vinduced.png | + | ====Step 5: Determining Vind==== |
- | ?440|Step 4 of the solution process. The sign of the V-induced is the **opposite** of the change | + | Now that the direction |
+ | $$V_{ind} = -\frac{d \Phi_B}{dt}$$ | ||
- | [{{ 184_notes: | + | For our example, |
- | ?440|Step 5 of the solution process. This allows | + | |
- | \\ | + | [{{184_notes: |
+ | ?440|Step 5: The sign of the V-induced is the **opposite** of the change in flux. }}] | ||
+ | ====Step 6: Determining the direction of Iind==== | ||
+ | Our final step then is to determine the direction of the induced current! This is where the right hand rule part comes in. But what does it mean for Vind to be positive or negative? Ultimately this goes back to our choice of d→A at the beginning. So there will be two scenarios: | ||
+ | * If Vind is positive, you should put the thumb of your right hand in the **same direction as your d→A**. Then the way that you curl your fingers will show the direction of the induced current. | ||
+ | * If Vind is negative, you should put the thumb of your right hand in the **opposite direction as your d→A**. Then the way that you curl your fingers will show the direction of the induced current. | ||
- | =====Concluding===== | + | For our example, we found that $V_{ind}$ was positive. So this means that we would stick our thumb in the -x direction |
- | $\frac{d \Phi_B}{dt}$, remember, is the change in flux over a given time. Meaning | + | |
- | \\ | + | |
- | \begin{align*} | + | |
- | -\int \vec{E}_{nc} \bullet d\vec{l} &= \frac{d \Phi_B}{dt} &&&& | + | |
- | -V_{ind} &= \frac{d \Phi_B}{dt} &&&& | + | |
- | \end{align*} | + | |
- | \\ | + | |
- | This is now a built in check to see if the predicted d→A was correct! __**If the V-induced value is positive, then your right hand will stay in the same direction | + | |
- | \\ | + | |
- | Remember that the thumb represents the $d\vec{A}$ and the curl of your fingers describes the rotational direction that the current will flow. In our example, since the d→A points in the positive x-direction | + | [{{184_notes: |
- | + | ?440|Step 6: Find the direction | |
+ | The right hand rule for determining the induced current direction is definitely more complicated than our previous right hand rules; however, this table helps break the process down into manageable chunks. We definitely recommend writing out each step in this way - otherwise it is easy to miss a step or a negative sign, which will throw off your final result. |