Problem

At what angular velocity in rpm will the peak vol…

03:03

Problem 28 Hard Difficulty

Calculate the peak voltage of a generator that rotates its 200-turn, 0.100 m diameter coil at 3600 rpm in a 0.800 T field.

Answer

474 $\mathrm{V}$

Top Physics 102 Electricity and Magnetism Educators

Christina K.

Rutgers, The State University of New Jersey

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Video Transcript

in this question. We have a generator that is rotating at 6 36 100 rpm and a 0.8 Kessler field. The diameter of the coil is 0.1 m and the number of turns is 200 were asked to calculate the peak voltage for this generator. So this is actually a relatively simple question when you consider the fact that we have a formula given in the text for the peak voltage of such a generator. So it just depends on these four variables and A B and omega. And the only thing that we need to be careful about here is just, you know, using the correct units, um, for this equation. So let's get started here. We're going to use this equation and I'm going to write out the Given. So N is the number of turns in the coil. Um, that's going to be 200. A. Should be the cross sectional area off the coil. So in the question, we're told that the diameter is 0.1 m. Um, that would make the radius 0.5 m, and we can use that to calculate the area that's going to mean the area is pi r squared or pie 0.5 m squared. So presumably the cross sectional area is a circle, and we can use that radius to calculate the area. So once we do that, we get an area of 7.85 times, 10 to the minus three meter squared. That's the area in the question. We were also given the strength of the magnetic field that was 0.8 Tesla's and then last but not least, we have the Omega, um, the frequency that's given as 3600 rotations permanent. But in order to use this correctly in the calculation, we do want this to be in radiance for second instead of rotations permanent. So let's go ahead and perform that conversion. So I'm going to write this as rotations per minute, and then I'm going to multiply by a conversion factor to take care of the minutes in two seconds. So I'm going to do one minute is equal to 60 seconds, so that takes care of the time, and then we want to convert this into radiance or two Pi. Radiance is equal to one rotation, so In order to correctly convert this, we've got Thio. Divide by 60 and multiply by two pi. Um, so that's going to give us 300 and 77 route, um radiance per second. So now that we have all of our givens written out, we can just use that original formula that I wrote down to calculate the peak MF. So the reason that it's peak is that the e m f will oscillate as the coils rotate through the magnetic field. So basically, the peak is gonna be the maximum that the IMF achieves. Okay, so we're just gonna go ahead and multiply that all together. And since we've used a standard units for everything, the final answer should be in volts. Okay, so rounding to three significant figures, the peak MF is going to be 474 Fox. So this is our final answer here

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