A 100-turn square coil of side 20.0 cm rotates about a vertical axis at ?= 1.50
ID: 1501751 • Letter: A
Question
A 100-turn square coil of side 20.0 cm rotates about a vertical axis at ?= 1.50 x 10^3 revolutions per minute. The horizontal component of the Earth’s magnetic field at the coil’s location is equal to 2.00 x 10^-5 T. (a) Calculate the maxium emf induced in the coil by this field. (b) What is the orientation of the coil with respect to the magnetic field (for instance, perpendicar to the field, or parallel to the field) when the maximum emf occurs? Experts only please and also please show all work and drawings and be clear! Please answer all parts!
Explanation / Answer
a) The induced EMF is given by the derivative of the magnetic flux x (-1). In this case, the flux is:
Flux = 100*A*B*cos(w*t), where:
A = 0.04 m^2
B = 2e-5 T
w = 1.5e3 rev/min=25 rev/s
So EMF = - (- 100*A*B*w) *sin(wt)
= 100*0.04*2e-5*25
= 200e-5 sin(wt)
The maximum is 2mV
b) orientation of the coil is perpendicar to the field.
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