..il T-Mobile Wi-Fi ? 9:47 PM 76% EXAMPLE 20.5 emf Induced in an AC Generator GO
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Question
..il T-Mobile Wi-Fi ? 9:47 PM 76% EXAMPLE 20.5 emf Induced in an AC Generator GOAL Understand physical aspects of an AC generator PROBLEM An AC generator consists of eight turns of wire, each having area A o.0900 m2, with a total resistance of 12.0 ?. The coil rotates in a magnetic field of 0.500 T at a constant frequency of 60.0 Hz, with axis of rotation perpendicular to the direction of the magnetic field. (a) Find the maximum induced emf. (b) What is the maximum induced current? (c) Determine the induced emf and current as functions of time. (d) What maximum torque must be applied to keep the coil turning? STRATEGY From the given frequency, calculate the angular frequency and substitute it, together with given quantities into the maximum emf equation. As functions of time, the emf and current have the form A sin ot, where A is the maximum emf or current, respectively. For part (d), calculate the magnetic torque on the coil when the current is at a maximum. The applied torque must do work against this magnetic torque to keep the coil turning (A) Find the maximum induced emf First, calculate the angular frequency 2ay . 2r (60.0 Hz) 377 rad/s . NABu, . 8(0.0900 m?)(0.500 TX377 rad/s) -136 V Substitute the values for N, A, B, and into the max emf equation, obtaining the maximum induced emf (B) What is the maximum induced current? 136 11.3A Substtute the maximum induced e Emau and the resistance R into Ohms law to find the maximum induced current 12.0 ? (C) Determine the induced emf and the current as functions of time E sin(136 v) sin 377 Substitute &u.nd; ? into the equation to the right to obtain the variation of ? with tirne , in seconds. The time variation of the current looks (11.3 A) sin 377 just ike this expression, except withExplanation / Answer
Question :
induced emf = 136*2
= 272 V (if frequency is diubled induced emf alos becomes double)
Practice it:
a) maximum induced emf, Vmax = N*A*B*w
= N*A*B*2*pi*f
= 8*0.092*0.412*2*pi*61.9
= 118 V
b) Imax = Vmax/R
= 118/13.3
= 8.87 A
c) V = Vmax*sin(w*t)
= Vmax*sin(2*pi*f*t)
= 118*sin(2*pi*61.9*t)
= 118*sin(389*t)
I = Imax*sin(w*t)
= 8.87*sin(389*t)
d) maximum torque = N*Imax*A*B
= 8*8.87* 0.92*0.412
= 26.9 N.m
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