The flexible loop in the figure below has a radius of 10.0 cm and is in a magnetic field of magnitude 0.160 T. The loop is grasped at points A and B and stretched until its area is nearly zero. If it takes 0.210 s to close the loop, what is the magnitude of the average induced emf in it during this time interval?

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The flexible loop in the figure below has a radius of 10.0 cm and is in a magnetic field of magnitude 0.160 T. The loop is grasped at points A and B and stretched until its area is nearly zero. If it takes 0.210 s to close the loop, what is the magnitude of the average induced emf in it during this time interval?
 mV

image
asked Feb 28, 2012 in Physics by Joey33 ~Expert~ (1,216 points)
    

1 Answer

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Best answer

According to Faraday's law of induction, emf induced in a loop is directly proportional to the time rate of change of magnetic flux through the loop:

so Emf = - change in magnetic flux / change in time

Magnetic Flux = integral of (magnetic field times change in Area)

So the change in magnetic flux is simply:

change in magnetic flux = Magnetic field * change in Area

 

Area = pi*r^2 

r = 10 cm = 0.1 m

Area = pi*(0.1 m)^2

Magnetic field = 0.16 T

t = 0.21 s

substituting into the formula for Emf we get:

Emf = (0.16 T) *(pi*(0.1 m)^2) / (0.21 s)  = 0.0239 V  = 23.9 mV

answered Mar 3, 2012 by kirby ~Expert~ (3,020 points)
selected Mar 3, 2012 by Joey33
Thanks! Great Answer.

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