Question

Your friend decides to generate electrical power by rotating a 100,000 turn coil of wire around an axis in the plane of the coil and through its center. The coil is perpendicular to Earth's magnetic field in a region where the field strength is equal to 0.300 G. The loops of the coil have a radius of 23 cm, and the coil has negligible resistance.
a. If your friend turns the coil at a rate of 140 rev/s, what peak current will exist in a 1400 O resistor that is connected across the terminals of the coil?
b. The average of the square of the current will equal half of the square of the peak current. What will be the average power delivered to the resistor?

Answer 1

Answer:

a) I=35mA

b) P=1.73W

Explanation:

a) The max emf obtained in a rotating coil of N turns is given by:

$emf_{max}=NBA\omega$

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

$A=\pi r^2 =\pi(0.23m)^2=0.16m^2$

$emf_{max}=(100000)(0.3*10^{-4}T)(0.166m^2)(140\frac{rev}{s})=69.72V$

Finally, the peak current is given by:

$I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA$

b)

we have that

$I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A$

$P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W$

hope this helps!!