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A 10.2 cm diameter wire coil is initially oriented so that its plane is perpendicular...

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lkehoe | Valedictorian

Posted April 14, 2013 at 4:37 PM via web

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A 10.2 cm diameter wire coil is initially oriented so that its plane is perpendicular to a magnetic field of 0.98 T pointing up.  During the course of 0.15 seconds, the field decreases to 0.10 T in the same direction.  What is the average induced emf in the coil?

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted April 15, 2013 at 5:35 AM (Answer #1)

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Remember we have the expressions:

`E = A * (dB)/(dt)`

`A = pi r^2`

`E = pi r^2 (dB)/(dt)`

where:

  • A = area of the coil
  • r = radius = diameter/2= 10.2/2 = 5.1 cm = 0.051 m
  • db = change in magnetic field = (0.98 - 0.10) T =
  • dt = time = 0.15 s

by plugging in the values, we can get the value of the average induced emf in the coil.

`E = pi r^2 (dB)/(dt)`

`E = pi (0.051)^2 (0.88)/(0.15)`

E = 0.04794 volt = 47. 9 mV -> answer

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