The rotating loop in an AC generator is a square 11.0 cm on a side. It is rotated at 60.0 Hz in a uniform field of 0.800 T. Calculate the following quantities as functions of time t, where t is in seconds.
(a) the flux through the loop
mT·m2
(b) the emf induced in the loop
V
(c) the current induced in the loop for a loop resistance of 2.00 Ω
A
(d) the power delivered to the loop
W
(e) the torque that must be exerted to rotate the loop
mN·m
2 answers
fdf
a) ABcos(wt)=(.11*.11)(.8)cos(2*pi*60t)
=.00968cos(377t)
b) ABwsin(wt)=(.11*.11)(.8)(2*pi*60)sin(2*pi*60t)=3.65sin(377t)
c) (ABw/R)sin(wt)=((.11*.11)(.8)(2*pi*60)/2)sin(2*pi*60t)=1.82sin(377t)
d)((ABw)^2/R)(sin(wt))^2=(((.11*.11)(.8)(2*pi*60))^2/2)(sin(2*pi*60t))^2=6.66sin(377t)
e)(A^2B^2w/R)(sin(wt))^2=(((.11*.11)(.8))^2*(2*pi*60)/2)(sin(2*pi*60t))^2=17.6sin(377t)
=.00968cos(377t)
b) ABwsin(wt)=(.11*.11)(.8)(2*pi*60)sin(2*pi*60t)=3.65sin(377t)
c) (ABw/R)sin(wt)=((.11*.11)(.8)(2*pi*60)/2)sin(2*pi*60t)=1.82sin(377t)
d)((ABw)^2/R)(sin(wt))^2=(((.11*.11)(.8)(2*pi*60))^2/2)(sin(2*pi*60t))^2=6.66sin(377t)
e)(A^2B^2w/R)(sin(wt))^2=(((.11*.11)(.8))^2*(2*pi*60)/2)(sin(2*pi*60t))^2=17.6sin(377t)