Problem

(a) If the emf of a coil rotating in a magnetic f…

18:40

Problem 36 Hard Difficulty

A 75-turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00 rad/s in a 1.25 T field, starting with the plane of the coil parallel to the field. (a) What is the peak emf? (b) At what time is the peak emf first reached? (c) At what time is the emf first at its most negative? (d) What is the period of the AC voltage output?

Answer

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Physics 102 Electricity and Magnetism

College Physics for AP® Courses

Chapter 23

Electromagnetic Induction, AC Circuits, and Electrical Technologies

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Video Transcript

diagram of the system. We know that. Then the instantaneous voltage would be equaling to the maximum voltage time sign of data. And we know that here Fada plus five is equaling 90 degrees and we can say that then the instantaneous voltage would be equaling the maximum voltage time sign of 90 degrees minus phi. Or we can say simply, this would be equaling to the maximum voltage times co sign of Omega Times teeth. And so for party to find the maximum voltage. This is simply gonna be equaling the number of turns times the area multiplied by Be the cross. Most of it maybe be the magnitude of magnetic field multiplied by omega, the angular velocity. Now, uh, we can simply solve this would be equaling, get into line number of turns. We know 75 then we have pi. Times are the radius of 0.500 meters quantity squared, multiplied by 1.25 10 slurs multiplied by the angular velocity of eight radiance per second. And we then find that the maximum voltage is 5.89 rounded to three significant figures. This would be our final answer for part a 5.89 volts. Now, for part B, we know that from the diagram we haven't instantaneous IMF equaling the maximum voltage times co sign of Omega T. And so, essentially here, four part B. We know that this is the current expression because, uh, the voltage IMF must change sign when fi is equaling omega T, which would equal pi over too. However, here we can say that the instantaneous M f equals the maximum e m f when co sign of Omega T equals one, or we can say omega T is equaling zero. And so Peak E M f um, occurs rather because it occurs AT T equals zero right after generator is turned on. So I would be your answer for part B t equals zero. And for part c, we know that the e m Memphis equaling the negative voltage essentially the negative maximum voltage when omega T is equaling pie. Because, of course, coast on a pious negative one. And so t would be equaling pie over omega. And so t is gonna be equaling pi divided by 8.0 radiance per second. And we find that tea is equaling 0.393 seconds. This would be our answer. Four parts. See where the instantaneous CMF is equaling two negative. The negative maximum, e m f and four part D. We know that the, uh, period is found by substituting omega into the period equation to pie over omega. So I would be the period, and this would be equaling two pies over 8.0 radiance per second. And we find that this is equal in 0.785 seconds. This would be our period. This would be our period for the system. That is the end of the solution. Thank you for watching.

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