Posted by Ulises on Tuesday, April 9, 2013 at 4:38pm.
A square loop (side L) spins with angular frequency ω in field of strength B. It is hooked to a load R.
Write an expression for current I(t) in terms of B,L,R,ω and t (enter omega for ω).
(b) How much work is done by the generator per revolution? Express your answer in terms of B,L,R and ω (enter omega for ω).
(c) To make it twice as hard to turn (twice as much work), what factor would you have to multiply the resistance R?
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physics  MeSelf, Wednesday, April 10, 2013 at 3:26am
a) L^2*B*omega*sin(omega*t)/R

physics  MeSelf, Wednesday, April 10, 2013 at 4:04am
the magnetic flux will be:
Φ = A*B cos(w*t)
Then the electromotrix force will be:
dΦ/dt = A*B*w*sin(w*t) = ε(t)
Since the current is given by
I(t) = ε(t)/R
we have
I(t) = L^2*B*w*sin(w*t)/R

physics  Espartan, Wednesday, April 10, 2013 at 8:15am
c)the factor is 1/2

physics  sat, Wednesday, April 10, 2013 at 2:24pm
part b plsssss
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