A thin square coil has 55.0 turns of conducting wire. It is rotating with consta

Question

A thin square coil has 55.0 turns of conducting wire. It is rotating with constant angular velocity in a uniform magnetic field B of 0.371 T. 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 the coil as function of time. (Note that the Flux is in thousandths of Webers (Wb). 1 Wb = 1T*m) (Also note that the first two answers require that you pull values off the graph. Once you get the answer correct, you should use the computers answer for subsequent calculations.)

1. What is the cross sectional area of the coil?

2. Calculate the angular velocity of the coil in rad/s

3. Evaluate the magnitude of the maximum induced voltage Emax in the coil.

4. The total electrical resistance of the coil is 3.51 ?. Calculate the maximum power dissipated.

Explanation / Answer

N=55

B=0.371 T

1)

flux=N*B*A

1.5*10^-3=55*0.351*A

==>

A=1.5*10^-3/(55*0.351)

A=7.77*10^-5 m^2

2)

from the graph

time period T=0.85 sec

w=2pi/T

=2pi/0.85

=7.392 rad/sec

3)

from tha graph,

emf v=N*change in flux/time

v=N*slope

v=55*(0-1*10^-3)/(1.25-1.15)

v=0.55 volt

4)

power p=v^2/R

=0.55^2/3.51

=0.0862 w

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