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Johns Hopkins University

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Problem 40

You need to make a generator for your bicycle light that will provide an alternating emf whose peak value is 4.2 $\mathrm{V}$ . The generator coil has 55 turns and rotates in a 0.040 -T magnitude

$\vec{B}$ ficld. If the coil rotates at 400 revolutions per sccond, what must the area of the coil be to develop this emf? Describe any problems with this design (if there are any).

Answer

7.6 $\mathrm{cm}^{2}$

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## Discussion

## Video Transcript

in this question, you're trying to create a generator that can provide an alternating E M f whose peak value is ah 4.2. Both so noticed. This is peak value, so the maximum and the generator coil has 55 turns. So in a sequel of 55 it rotates in a constant magnetic field that has the strength 0.0 for zero Tesla, the coil rotates at 400 revolutions per second. So whenever you see something per second that is talking about frequency, so you have a frequency of 400 revs per second. One must be What must the area of the coil be to develop this CMF. So from this question, it seems like we're still going to use Faraday's law of magnetic induction. But the only difference here issue are given frequency. Um, noticed that the generator coil is rotating within a constant magnetic field. So what is changing? The flux in this case is the change in angle but were not given any angle. However, we can relate frequency to angle by using this relationship, we note that the angular velocity is equal to two pi times frequency. So if I plug in the frequency value of 400 here I find that Omega is secret of 2513 one over seconds. So this will help us because we understand that magnetic flux is equal to be a Times co sine of the angle. Now, when we're not given angle, but we have an angular velocity. We can rewrite angle fada as Hello Zain of Omega Times Time because given a, um, coil that rotates at constant speed constant, angular speed omega, then you're changing angle is actually equal to Omega Times the changing time. So we'll believe this part makes sense to you. Now let's translate that into Faraday's law of magnetic induction. So they're days law. Magnetic induction still starts out as end times change in magnetic flux over changing time. And, of course, I can go ahead and plug this part in four plus. And here we have this and the change and hopes flux over the change in time. So it is hard to, uh, talk about the change in the angle over the changing time, using this expression without going into you calculus. So I'm just going to tell you what the result of this will be, and it's already part of your textbook, so you can go back to the section that you read it for yourself as well. So this comes out to equal to this expression and be a times Omega Times sign, Ah, Omega T. And because we are looking at peak value, then we definitely want Sign of Omega Times, Time to equal toe one so that we don't have to worry about a value that is less than the maximum. In other words, we are left with Yeh Memphis equal to M B A Times omega. Now, this question becomes a lot easier because all you have to do in solving for a is to do some algebraic manipulation and plugging in and self. That's exactly what I'm doing right here and now. I am substituting the given values and is 55 the U 0.40 times omega, which is 25 13 and your area turns out to be 7.6 times 10 to the negative fourth meters squared. Make sure you have the right unit for area

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