A generator uses a coil that has 80 turns and a 0.50 T magnetic field. The frequency of this generator is 60.0 Hz, and its emf has an rms value of 120 V. Assuming that each turn of the coil is a square (an approximation), determine the length of the wire from which the coil is made.
I used B= ( sq rt 2 * Emf) / ( NA2piF)
I solved this for A
then I tried A=pir^2
I tried A=L*W
Nothing is working to give me the length of the wire can someone please explain pleae
1. June 2010 at 11:06 pm
emf = N*B*omega*A*sin(omega*t)
A = emf / [N*B*omega*sin(omega*t)]
emf = 120V * sqrt2 = 169.7V
omega = 2*pi*f = 377 rad/s
A = L^2 since it’s square, however, the length of one turn is 4L.
A = 169.7 / [80*0.5*377*sin(377*t)]
= 169.7 / [80*0.5*377] max val of sin = 1
A = 0.1125m
L = sqrt A = (0.1125m)^1/2 = 0.1061m
Length of one turn = 4*0.1061m = 0.4243m
Length of 80 turns = 80*0.4243m = 33.95m