The drawing shows a plot of the output emf of a generator as a function of time t.
The coil of this device has a cross-sectional area per turn of 0.020 m2 and contains 145 turns.
A.) Find the frequency f of the generator in hertz
B.) the angular speed ω in rad /s
C.) and the magnitude of the magnetic field
Please help! THis is from my online independent study of physics AP, so I don’t have a teacher. Please just help me understand this at least.
PICTURE!!!
http:// i39.tinypic .com/2uzfleu .gif
I don’t know why it wont let me submit this, so I put a spaces within the url and so please just copy and paste the entire site into another tab without any spaces.
Thanks =)
PICTURE!!!
http:// i39.tinypic .com/2uzfleu .gif
I don’t know why it would not let me submit this, so I put spaces within the url. Please just copy and paste the entire site into another tab without any spaces.
Thanks =)
9. July 2010 at 1:30 am
A) Frequency is the number of cycles per second. From the graph there is 1 cycle per .42 sec, frequency = 1/.42 = 2.38 cps
B) 2,38 cycles per sec corresponds to the generator rotating thru (2.38) x (2pi) radians per sec
w = 2(pi)f = (2)(3.14)(2.38) = 14.9 rad/s
C) The emf induced in a coil of N turns with changing rate of flux is;
E = -Nd[BACos(wt)]/dt , where wt is the angle between the direction of B and the direction of the normal to A.
In this case B & A are constant and the derivative is just ;
d[BACos(wt)]/dt = -BAwSin(wt)
Then the emf is;
E = NBAwSin(wt)
The max emf is then, Emax = NBAw
and; B = Emax/NAw
From graph, Emax = 28 volts
Also; A = .02
N = 145
w = 14.9
So; B = 28/(145)(.02)(14.9) = .65 T