184_notes:induced_current

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184_notes:induced_current [2021/11/12 23:01] – [Finding the Induced Current Direction] stumptyl184_notes:induced_current [2021/11/12 23:14] – [Step 5: Determining $V_{ind}$] stumptyl
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 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. Since our coil is next to the south side of the magnet, this means the magnetic field inside the coil will mostly be pointing to the left (in towards the south side of the magnet). So in the second column we will put an arrow to the left. 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. Since our coil is next to the south side of the magnet, this means the magnetic field inside the coil will mostly be pointing to the left (in towards the south side of the magnet). So in the second column we will put an arrow to the left.
  
-[{{184_notes:ic_bfield.png?440| Step 2: isolates the direction of the magnetic field and now places that corresponding vector into the chart. }}]+[{{184_notes:inductionchart_partb.png?440| Step 2: isolates the direction of the magnetic field and now places that corresponding vector into the chart. }}]
  
  
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 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't matter which way you pick, as long as the $d\vec{A}$ is perpendicular to the area. For this example, we'll pick the $d\vec{A}$ to point to the left, so we draw an arrow in the third column that points to the left. 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't matter which way you pick, as long as the $d\vec{A}$ is perpendicular to the area. For this example, we'll pick the $d\vec{A}$ to point to the left, so we draw an arrow in the third column that points to the left.
  
-[{{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.  }}] 
  
 ====Step 4.) $\Phi_{B,i}$, $\Phi_{B,f}$ and $\frac{d\Phi_{B}}{dt}$==== ====Step 4.) $\Phi_{B,i}$, $\Phi_{B,f}$ and $\frac{d\Phi_{B}}{dt}$====
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 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.
  
-[{{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 $V_{ind}$ column. For our example, the change in flux was negative. So we write down "positive" for the $V_{ind}$ column.
  
-[{{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.  }}]
  
  • 184_notes/induced_current.txt
  • Last modified: 2021/11/12 23:15
  • by stumptyl