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184_notes:induced_current [2021/11/12 23:13] – [Step 2.) Draw the direction of the B-field through the relevant area] stumptyl | 184_notes:induced_current [2021/11/12 23:15] (current) – [Step 1.) Draw a picture of your situation] stumptyl |
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{{youtube>Qlg0Iu1Do94?large}} | {{youtube>Qlg0Iu1Do94?large}} |
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| **NOTE: In this chart there is a mistake in the chart being used: The variables "V" and "I" are not vectors. Please make a note of this when watching the video to ensure there is no confusion.** |
==== 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/south), and which way the magnet is moving (marked with the velocity →v arrow). | 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/south), and which way the magnet is moving (marked with the velocity →v arrow). |
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[{{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:inductionchart_updated_11_12_2021.png?440| Step 1: First make a diagram of the particular situation include the coil, magnet, and relevant directions. |
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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 "out of” the coil. For our set up, this means that d→A could point either to the left or right (-x or +x direction). It doesn't matter which way you pick, as long as the d→A is perpendicular to the area. For this example, we'll pick the d→A to point to the left, so we draw an arrow in the third column that points to the left. | 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 "out of” the coil. For our set up, this means that d→A could point either to the left or right (-x or +x direction). It doesn't matter which way you pick, as long as the d→A is perpendicular to the area. For this example, we'll pick the d→A to point to the left, so we draw an arrow in the third column that points to the left. |
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[{{184_notes:ic_da.png?440| Step 3: Pick a direction for dA that points perpendicular to the coil. In this example, we pick dA to be to the left. }}] | [{{184_notes:inductionchart_partc.png?440| Step 3: Pick a direction for dA that points perpendicular to the coil. In this example, we pick dA to be to the left. }}] |
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====Step 4.) ΦB,i, ΦB,f and dΦBdt==== | ====Step 4.) ΦB,i, ΦB,f and dΦBdt==== |
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 to help you think through this. For example, we could take 2−10=−8.) So we write down in the sixth column that the change in flux is negative. | 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 to help you think through this. For example, we could take 2−10=−8.) So we write down in the sixth column that the change in flux is negative. |
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[{{184_notes:ic_flux.png | [{{184_notes:inductionchart_partd.png |
?440|Step 4: Determine the sign of the change in flux based on the initial and final flux for the situation. }}] | ?440|Step 4: Determine the sign of the change in flux based on the initial and final flux for the situation. }}] |
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For our example, the change in flux was negative. So we write down "positive" for the Vind column. | For our example, the change in flux was negative. So we write down "positive" for the Vind column. |
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[{{184_notes:ic_vinduced.png | [{{184_notes:inductionchart_parte.png |
?440|Step 5: The sign of the V-induced is the **opposite** of the change in flux. }}] | ?440|Step 5: The sign of the V-induced is the **opposite** of the change in flux. }}] |
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For our example, we found that Vind 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 of the coil. Just because it's often hard to draw the current direction in 3D, we often will just notate where on the coil the current is into/out of the page. | For our example, we found that Vind 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 of the coil. Just because it's often hard to draw the current direction in 3D, we often will just notate where on the coil the current is into/out of the page. |
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[{{184_notes:inductionrhr.png | [{{184_notes:inductionchart_partf.png |
?440|Step 6: Find the direction of the induced current based on the right hand rule. }}] | ?440|Step 6: Find the direction of the induced current based on the right hand rule. }}] |
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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. | 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. |