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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/potential, and how that applies to voltage transformers.
Alternating Current
Add Figure 14.6
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 square waves) or triangular shapes (or triangle waves). We will stick with a sine graph because it is mathematically easier to manipulate and is the most common kind of alternating current. This means that we represent the current as: $$I=I_0sin(2\pi \cdot f \cdot t)$$ 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, frequency tells you how much of a full cycle happens in 1 second. For example, if your frequency is $1 Hz$, this means that your current goes from the maximum current to the minimum current and back to the maximum current in 1 second. If your frequency is $0.5 Hz$, your current would go from the maximum current to the minimum current (or half of a cycle) in one second. In short, frequency tells you what fraction of a complete cycle or how many complete cycles occur in 1 second.
Alternatively, we could talk about the period of the oscillation, which tells you how many seconds it takes to complete one full cycle and is inversely related to frequency: $$\tau = \frac{1}{f}$$ This means that a larger period would be related to a smaller frequency, and a smaller period would be related to a higher frequency. Since the period is often easier to think about conceptually, it may be easier to start with period (from a graph for example) and then relate it back to the frequency.
Voltage Transformer
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/current in any nearby loop of wire since the magnetic field would be constantly changing. One common application of this idea is called a step-up transformer (or alternatively a step-down transformer), which is crucial to getting electricity from the power generator to your house.
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
For this problem it is easy to overcomplicate the problem and get very lost in the mathematics if you do not think about the problem conceptually first. I would highly suggest having your groups think conceptually about what would happen in this “basic transformer” and then write out the math steps that reflect those steps.
Step 1: There is an oscillating potential/current in the first solenoid. This will create a constantly changing magnetic field. This means we can write the potential as a sin function (though the specific equation is not particularly needed). $$V_1=V_0 sin(\omega t)$$
Step 2: If you pick a material (like iron) that easily responds to the magnetic field, then this changing magnetic field will exist everywhere in the cylinder. For this solution, we will assume that the material is iron and assume the cylinder has a constant area. This tells us that: $$B_1=B_2$$ $$A_1=A_2$$ Note that $B_2$ here is the magnetic field that exists in the second solenoid by being propagated through the iron from the first solenoid. (It is not due to a current in the second solenoid.)
Step 3: If we put the second solenoid on the end of the bar, this changing magnetic field (from the first solenoid) will induce a potential in the second solenoid. We can use Faraday's Law to write this as: $$-V_{2}=\frac{d\Phi_{B2}}{dt}$$
Step 4: We can rewrite the $\Phi_{B2}$ using the flux definition: $$\Phi_{B2}=\int \vec{B}_2 \bullet \vec{dA}_2$$ The direction of the magnetic field would always be perpendicular to the area of the cylinder - so this simplfies to: $$\Phi_{B2}=B_2A_2$$ But this is the magnetic flux through one loop in the solenoid - if we want the total flux through all the loops then we have to multiply by the number of loops: $$\Phi_{B2}=B_2A_2N_2$$ 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 to time (not add/taking away loops or increasing/decreasing the area), so those can come out of the derivative: $$-V_{2}=N_2A_2\frac{d}{dt}\biggl(B_2\biggr)$$ And we can use what we found from Step 2 to rewrite the flux in terms of source B-field and the source area: $$-V_{2}=N_2A_1\frac{d}{dt}\biggl(B_1\biggr)$$
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: $$\frac{d\Phi_{B1}}{dt}=\frac{d}{dt}\biggl(B_1 A_1\biggr)$$ If we use Faraday's Law, we can rewrite the flux through a single loop in terms of the number of coils and the induced voltage in the whole solenoid : $$-V_1=N_1 \frac{d}{dt}\biggl(B_1 A_1\biggr)$$ $$\frac{d}{dt}\biggl(B_1 A_1\biggr)=\frac{-V_1}{N_1}$$
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: $$-V_2=N_2\frac{-V_1}{N_1}$$ $$V_2=\frac{V_1 N_2}{N_1}$$
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!