A 117-turn square coil of side 20.0 cm rotates about a vertical axis at = 1.77 1
ID: 1452264 • Letter: A
Question
A 117-turn square coil of side 20.0 cm rotates about a vertical axis at = 1.77 103 rev/min as indicated in the figure below. The horizontal component of Earth's magnetic field at the coil's location is equal to 2.00 10-5 T.
(a) Calculate the maximum emf induced in the coil by this field. ans mV
(b) What is the orientation of the coil with respect to the magnetic field when the maximum emf occurs?
- The plane of the coil is parallel to the magnetic field.
- The plane of the coil is oriented 45° with respect to the magnetic field.
- The plane of the coil is perpendicular to the magnetic field.
please reduce anwser by unit given in the anwser blank.
Explanation / Answer
emf = NABw*sin wt
for maximum emf induced
sin wt = 1
EMF = NABw
w = 1.77*10^3 rev/min = 185.35 rad/sec
N = 117 turns
B = 2*10^-5 T
A = L^2 = (0.2)^2 = 4*10^-2 m^2
EMF = 117*0.04*2*10^-5*185.35 = 0.0173 V
B.
EMF will be maximum when the plane of the coil is parallel to the magntic field, because we need sin theta = 1, and theta is the angle between magnetic field and normal of the loop.
The plane of the coil is parallel to the magnetic field.
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