A thin square coil has 39 turns of conducting wire. It is rotating with constant
ID: 1376207 • Letter: A
Question
A thin square coil has 39 turns of conducting wire. It is rotating with constant angular velocity in a uniform magnetic field B of 0.330T. At times there is NO magnetic flux through the coil, and at other times, there is the maximum possible flux. The graph below shows the magnetic flux PHI through ONE turn of the coil as a function of time.
What is the length of a side of the coil?
Calculate the angular velocity of the coil.
Evaluate the magnitude of the induced voltage in the coil, Vemf, at time t= 2.60 s.
Now connect the ends of the wire together to make a circuit. The electrical resistance of the coil is 3.14Ohm. Calculate the power dissipated at time t= 2.60s.
A thin square coil has 39 turns of conducting wire. It is rotating with constant angular velocity in a uniform magnetic field B of 0.330T. At times there is NO magnetic flux through the coil, and at other times, there is the maximum possible flux. The graph below shows the magnetic flux PHI through ONE turn of the coil as a function of time. What is the length of a side of the coil? Calculate the angular velocity of the coil. Evaluate the magnitude of the induced voltage in the coil, Vemf, at time t= 2.60 s. Now connect the ends of the wire together to make a circuit. The electrical resistance of the coil is 3.14Ohm. Calculate the power dissipated at time t= 2.60s.Explanation / Answer
The maximum flux is
Bfluxmax = N B s^2
--> s = sqrt [Bfluxmax / (N B)]
as Bfluxmax = 1.5E-3 T m^2,
s = 0.0107 m [ANSWER, side length]
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As we can see, the period oscillation from the graph is
T = 0.80 s
Thus, as
w = 2 pi / T
then
w = 7.8 rad/s [ANSWER, angular velocity]
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