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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 - ,

    a) L^2*B*omega*sin(omega*t)/R

  • physics - ,

    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 - ,

    c)the factor is 1/2

  • physics - ,

    part b plsssss

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