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184_notes:induced_current [2021/04/05 20:44] – [Concluding] stumptyl | 184_notes:induced_current [2021/11/12 23:14] – [Step 6: Determining the direction of $I_{ind}$] stumptyl | ||
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- | =====Induced Current | + | =====Finding the Induced Current |
- | Now that we have taken some time to discuss | + | So far, we've been talking about the different pieces of Faraday' |
- | [{{ 184_notes: | + | $$V_{ind} = -\frac{d \Phi_B}{dt}$$ |
+ | 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: | ||
- | ====Right Hand Rule==== | + | This video will walk you through an example of how to use this table or you can read about it in the notes below. |
- | \\ | + | |
- | As we have learned from previous notes, when discussing the right-hand rule for the induced current the focus is on the $d\vec{A}$. In your right hand, make a fist and place the thumb out. The thumb in this situation will be representing your $d\vec{A}$. The fist itself will be used to predict the induced current flow. Normally, these situations could be described as clockwise and counterclockwise when describing the direction of flow; however, as you may notice that by simply saying “Counter-clockwise” another observer would still be confused because this statement involves assumptions about the observation point. By utilizing this Right Hand Rule model, you will be able to grasp a better understanding of the direction of flow of the Induced Current as well as be able to articulate the direction to the best of your abilities also. | + | {{youtube> |
- | \\ | + | ==== Right Hand Rule Steps ==== |
+ | ==== Step 1.) Draw a picture of your situation ==== | ||
+ | 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: | + | [{{184_notes: |
+ | }}] | ||
- | \\ | ||
- | =====Approach====== | + | ====Step 2.) Draw the direction of the B-field through the relevant area==== |
- | Let's take a moment and walk through how to solve the following | + | Next we need to determine |
- | \\ | + | [{{184_notes: |
- | [{{ 184_notes: | ||
- | \\ | ||
+ | ====Step 3.) Choose the $d\vec{A}$ ==== | ||
+ | Remember that the $d\vec{A}$ is perpendicular to the cross section area of the coils. Meaning, that you can think of the $d\vec{A}$ as pointing "out of” the coil. For our set up, this means that $d\vec{A}$ could point either to the left or right (-x or +x direction). It doesn' | ||
- | ====Step 1.) Direction of B-field==== | + | [{{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. | + | |
+ | ====Step 4.) $\Phi_{B, | ||
+ | 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? | ||
- | \\ | + | The magnitude |
- | [{{ 184_notes: | + | $$\Phi_B = \int \vec{B} \bullet d\vec{A} = \int B *dA *cos(\theta)$$ |
- | \\ | + | So if $\theta$ is between $0^\circ-90^\circ$, |
- | ====Step 2.) Choose the $d\vec{A}$ ==== | + | ===Initial flux=== |
- | \\ | + | The initial flux is determined for the " |
- | Remember that the $d\vec{A}$ | + | |
- | \\ | + | ===Final Flux=== |
- | [{{ 184_notes: | + | The final flux is determined after some time has passed. Think about this as looking at the “End |
- | ?440|Step 3 of the solution process. This step takes the movement of the object and generalizes | + | |
- | \\ | + | |
+ | ===Change in Flux=== | ||
+ | Now we can determine the sign of the change in flux. Since we already have the initial and final flux values, this step is simple. We just use: | ||
+ | $$\Delta \Phi_B = \Phi_{B,f} - \Phi_{B, | ||
- | ====Step 3.) $\frac{d \Phi_B}{dt}$==== | + | In our case, this means we'd be taking a small positive number minus a big positive number. This will result in a //negative// change in flux. (If it helps, you can assign numbers |
- | \\ | + | |
- | This portion of the approach is where the generalization of the situation begins. Meaning now we are going to talk about what the items: Big/small and Positive/Negative will mean. When we refer to Big and Small, we are using these terms to understand the relative impact that the situation has before and after with whatever object is moving. As we've seen before, | + | |
- | When referring to positive/ | + | [{{184_notes: |
+ | ?440|Step 4: Determine | ||
- | \\ | + | ====Step 5: Determining $V_{ind}$==== |
+ | Now that the direction of the $\frac{d | ||
+ | $$V_{ind} = -\frac{d \Phi_B}{dt}$$ | ||
- | ====Initial flux==== | + | For our example, the change in flux was negative. So we write down " |
- | Now that we have an understanding of the key terms we can start to walk through the table. The initial flux will be done __BEFORE THE POSITION__ CHANGE of the object. Think about this as looking at the “Beginning of the Scenario”. From this observation, consider how much of an impact | + | |
- | \\ | + | |
- | ====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”. | + | |
+ | [{{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 $I_{ind}$==== | ||
+ | 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 $V_{ind}$ to be positive or negative? Ultimately this goes back to our choice of $d\vec{A}$ at the beginning. So there will be two scenarios: | ||
+ | * If $V_{ind}$ is positive, you should put the thumb of your right hand in the **same direction as your $d\vec{A}$**. Then the way that you curl your fingers will show the direction of the induced current. | ||
+ | * If $V_{ind}$ is negative, you should put the thumb of your right hand in the **opposite direction as your $d\vec{A}$**. Then the way that you curl your fingers will show the direction of the induced current. | ||
- | [{{ 184_notes: | + | For our example, we found that $V_{ind}$ was positive. So this means that we would stick our thumb in the -x direction (pointing to the left) and then curl our fingers around in a circle. You should find the the current would go into the page at the top of the coil and would come back out of the page at the bottom |
- | ?440|Step 4 of the solution process. The sign of the V-induced is the **opposite** | + | |
- | [{{ 184_notes:inducedcurrentscenario_final.png | + | [{{184_notes: |
- | ? | + | ? |
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- | \\ | + | |
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- | + | ||
- | =====Concluding===== | + | |
- | $\frac{d \Phi_B}{dt}$, | + | |
- | \\ | + | |
- | \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 $d\vec{A}$ was correct! __**If the V-induced value is positive, then your right hand will stay in the same direction that you started with. If the V-induced value is negative, then you must flip your hand to go in the opposite direction.**__ Now that we have the properly identified $d\vec{A}$, we can utilize the Right Hand Rule Method to describe the direction of the I-induced. | + | |
- | \\ | + | |
- | + | ||
- | 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 $d\vec{A}$ points in the positive x-direction and the V-induced was positive, Meaning that if the induced current was occurring in the z-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. |