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184_notes:ac [2017/11/24 16:41] – dmcpadden | 184_notes:ac [2018/08/09 19:21] – [Voltage Transformer] curdemma | ||
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+ | Section 22.2 in Matter and Interactions (4th edition) | ||
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===== Changing Flux from an Alternating Current ===== | ===== Changing Flux from an Alternating Current ===== | ||
As we said before, one of the most important sources of a changing magnetic field is an alternating current. This is what actually comes out of the wall outlets; as opposed to the current from a battery which is a constant current (or a direct current). We are only briefly going to talk about alternating current as it refers to induction and changing magnetic flux, but there are many more applications of alternating current, especially with regard to circuits, resistors, and capacitors. For the purposes of these notes, we will talk about how we represent an alternating current, how that alternating current can produce an induced current/ | As we said before, one of the most important sources of a changing magnetic field is an alternating current. This is what actually comes out of the wall outlets; as opposed to the current from a battery which is a constant current (or a direct current). We are only briefly going to talk about alternating current as it refers to induction and changing magnetic flux, but there are many more applications of alternating current, especially with regard to circuits, resistors, and capacitors. For the purposes of these notes, we will talk about how we represent an alternating current, how that alternating current can produce an induced current/ | ||
==== Alternating Current ==== | ==== Alternating Current ==== | ||
- | FIXME Add Figure 14.6 | + | [{{ 184_notes: |
When a current is constantly switching between positive and negative values, we call this an **alternating current**. Typically, this means that we have a current that is represented by a sine graph (shown in the figure to the right), though there can be other types of alternating current that look like step functions (also called [[https:// | When a current is constantly switching between positive and negative values, we call this an **alternating current**. Typically, this means that we have a current that is represented by a sine graph (shown in the figure to the right), though there can be other types of alternating current that look like step functions (also called [[https:// | ||
- | $$I=I_0sin(2\pi \cdot f \cdot t)$$ | + | $$I=I_0\sin(2\pi \cdot f \cdot t)$$ |
where $I$ is the current at any particular time $t$, $I_0$ is the //maximum// current (this is a constant value), and $f$ is the frequency, which in general tells you how fast the current is changing between positive and negative with units of $\frac{1}{s}$ or $Hz$ (Hertz). More specifically, | where $I$ is the current at any particular time $t$, $I_0$ is the //maximum// current (this is a constant value), and $f$ is the frequency, which in general tells you how fast the current is changing between positive and negative with units of $\frac{1}{s}$ or $Hz$ (Hertz). More specifically, | ||
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If you have an oscillating current, this would also mean that you would have an oscillating magnetic field everywhere around the wire (since currents create magnetic fields). If there is an oscillating magnetic field, this means that there will also be an induced potential/ | If you have an oscillating current, this would also mean that you would have an oscillating magnetic field everywhere around the wire (since currents create magnetic fields). If there is an oscillating magnetic field, this means that there will also be an induced potential/ | ||
- | As you figured out in the project last week, a rotating loop in a magnetic field will create a oscillating current and oscillating voltage. This idea is exactly what happens in a power generator, just on a much larger scale. In a power generator there are thousands of loops, rotating very quickly in a large magnetic field to produce a very large current with a frequency of 60 Hz (in the U.S. at least). However, because power (or the energy transferred to heat) is related to $P=I^2R$, sending a very large current over large distances (like from the generator to your house) would result in a large loss of power over that wire (). This is a problem because any power lost on | + | [{{184_notes: |
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+ | As you figured out in the project last week, a rotating loop in a magnetic field will create a oscillating current and oscillating voltage. This idea is exactly what happens in a power generator, just on a much larger scale. In a power generator there are thousands of loops, rotating very quickly in a large magnetic field to produce a very large current with a frequency of 60 Hz (in the U.S. at least). However, because power (or the energy transferred to heat) is related to $P=I^2R$, sending a very large current over large distances (like from the generator to your house) would result in a large loss of power over that wire. This is a problem because any power lost on the wires can't be used in your house and if that power is large enough it could heat up the wires to point where they are damaged. | ||
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+ | To get around this problem, a step up transformer is used to change a low voltage, high current circuit (like what comes out the generator) into a high voltage, low current circuit for transport from the generator to the neighborhoods or wherever it is needed. A step down transformer is then used close to the neighborhoods to return the high voltage, low current back to a low voltage, high current circuit that is then used in your house. You may have seen these around your neighborhood - they look like small boxes attached to the power lines overhead, generally on the lines going from a larger street into a residential area (shown in the figure to the left). | ||
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+ | [{{ 184_notes: | ||
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+ | In these notes, we will go through how a step up transformer works and how it uses induction to change the voltage from a low voltage to a high voltage. We will use a basic transformer, | ||
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+ | [{{184_notes: | ||
- | For this problem it is easy to overcomplicate | + | Because there is an oscillating potential/ |
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+ | [{{ 184_notes: | ||
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+ | If we put the secondary solenoid on the end of the iron ring, this changing magnetic field will be the same as that from the primary solenoid: $B_P=B_S$. This changing magnetic field (from the primary solenoid) will induce a voltage ($V_S$) in the secondary solenoid. We can use Faraday' | ||
+ | $$-V_{S}=\frac{d\Phi_{B_{S}}}{dt}$$ | ||
- | **Step 1**: There is an oscillating potential/ | + | We can then rewrite the flux $\Phi_{B2}$ using the flux definition: |
- | $$V_1=V_0 sin(\omega t)$$ | + | $$\Phi_{B_{S}}=\int \vec{B}_S \bullet \vec{dA}_S$$ |
+ | where the area here would be the cross-sectional area of the iron ring (since this is where the magnetic field is). The direction of the magnetic field would always be perpendicular to the area of the cylinder - so the flux would simplfy to: | ||
+ | $$\Phi_{B_{S}}=B_S A_S$$ | ||
+ | But this is the magnetic flux through one loop in the secondary solenoid - if we want the total flux through all the loops then we have to multiply by the number of loops: | ||
+ | $$\Phi_{B_{S}}=B_S A_S N_S$$ | ||
+ | We can then plug this into the voltage | ||
+ | $$-V_{S}=\frac{d}{dt}\biggl(B_S A_S N_S\biggr)$$ | ||
- | **Step 2**: If you pick a material | + | Since $N_S$ and $A_S$ are constant with respect to time (not add/taking away loops or increasing/ |
- | $$B_1=B_2$$ | + | $$-V_{S}=N_S A_S \frac{d}{dt}\biggl(B_S\biggr)$$ |
- | $$A_1=A_2$$ | + | //__If we assume |
- | Note that $B_2$ here is the magnetic field that exists in the second | + | $$-V_{S}=N_S A_P \frac{d}{dt}\biggl(B_P\biggr)$$ |
- | **Step 3**: If we put the second solenoid on the end of the bar, this changing magnetic | + | If we look at the $A_P\frac{d}{dt}\biggl(B_P\biggr)$ term, this looks very much like the changing magnetic |
- | $$-V_{2}=\frac{d\Phi_{B2}}{dt}$$ | + | $$\frac{d\Phi_{B_{P}}}{dt}=\frac{d}{dt}\biggl(B_P A_P\biggr)$$ |
+ | If we use Faraday' | ||
+ | $$-V_P=N_P \frac{d}{dt}\biggl(B_P A_P\biggr)$$ | ||
+ | $$\frac{d}{dt}\biggl(B_P A_P\biggr)=\frac{-V_P}{N_P}$$ | ||
- | **Step 4**: We can rewrite | + | Finally, |
- | $$\Phi_{B2}=\int \vec{B}_2 \bullet \vec{dA}_2$$ | + | $$-V_S = N_S \frac{-V_P}{N_P}$$ |
- | The direction of the magnetic field would always be perpendicular | + | $$V_S=V_P \frac{N_S}{N_P}$$ |
- | $$\Phi_{B2}=B_2A_2$$ | + | |
- | But this is the magnetic flux through one loop in the solenoid | + | |
- | $$\Phi_{B2}=B_2A_2N_2$$ | + | |
- | We can then plug this into the equation we wrote in Step 2: | + | |
- | $$-V_{2}=\frac{d}{dt}\biggl(B_2A_2N_2\biggr)$$ | + | |
- | **Step 5**: $N_2$ and $A_2$ are constant with respect | + | This equation then tells us that that if we want the voltage in the secondary solenoid |
- | $$-V_{2}=N_2A_2\frac{d}{dt}\biggl(B_2\biggr)$$ | + | |
- | And we can use what we found from Step 2 to rewrite | + | |
- | $$-V_{2}=N_2A_1\frac{d}{dt}\biggl(B_1\biggr)$$ | + | |
- | **Step 6**: If we look at the $A_1\frac{d}{dt}\biggl(B_1\biggr)$ term, this looks very much like the changing magnetic flux (or induced voltage) through a single loop of the //first// solenoid: | + | ==== Examples ==== |
- | $$\frac{d\Phi_{B1}}{dt}=\frac{d}{dt}\biggl(B_1 A_1\biggr)$$ | + | [[: |
- | If we use Faraday' | + | |
- | $$-V_1=N_1 \frac{d}{dt}\biggl(B_1 A_1\biggr)$$ | + | |
- | $$\frac{d}{dt}\biggl(B_1 A_1\biggr)=\frac{-V_1}{N_1}$$ | + | |
- | **Step 7**: We can plug this result into the equation from Step 5 then to get the potential in the second solenoid as it relates to the potential in the first solenoid: | + | [[:184_notes: |
- | $$-V_2=N_2\frac{-V_1}{N_1}$$ | + | |
- | $$V_2=\frac{V_1 N_2}{N_1}$$ | + | |
- | Since we want an increase in potential by a factor of 10, this equation then tells that the second solenoid needs to have a 10 times as many coils as the first solenoid. So they could pick either the 10 and 100 turn solenoids or the 20 and 200 turn solenoids. Note, you do not really need to do any derivatives or flux calculations in this problem! |