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?

No one has answered this question yet.

- 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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