Recitation Group Problem #13 — Solutions
Department of Physics PHYS 408
University of New Hampshire May 1, 2003
The Electric Generator
1. the EMF generated in the circuit, as a function of time: The area of the coil is: A=w `.
The flux through the coil is:
Φ(t) = N A B sin ωt ,
so the EMF is given by:
E(t) = dΦ(t)
dt=N A B ω cos ωt
The answer could also be sin ωt, depending on orientation at t= 0.
2. the current through the resistor as a function of time,
i(t) = E(t)
R=N A B ω
Rcos ωt .
3. the power lost in the resistor as a function of time,
P(t) = i2(t)R=(NAB ω)2
Rcos2ωt . (1)
4. the average power lost in the resistor over a cycle,
hPi=1
TZT
0
P(t) dt=(NAB ω)2
2R.
5. The force on each length of the coil is:
F(t) = N B` i(t) = N2A` B 2ω
Rcos ωt
so the torque on the coil is:
τ(t) = 2 F(t)w
2cos ωt =N2A2B2ω
Rcos2ωt .
6. the power provided by the external torque to turn the generator, as a function of time
is:
P(t) = ω τ(t) = (N AB ω)2
2Rcos2ωt , (2)
which is the same as the power lost in the resistor in Eq. (1). Energy has been
conserved.
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