A 32-turn circular coil of wire has diameter 1.10 m. It is placed with its axis along the direction of the Earth's magnetic field of 47.0 µT and then in 0.200 s is flipped 180°. An average emf of what magnitude is generated in the coil?

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32-turn circular coil of wire has diameter 1.10 m. It is placed with its axis along the direction of the Earth's magnetic field of 47.0 µT and then in 0.200 s is flipped 180°. An average emf of what magnitude is generated in the coil?


 mV

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 = - change of the magnetic flux with respect to time

If a coil consists of N loops with the same area and magnetic flux, an emf is induced in every loop, and so the total induced emf becomes:

Suppose the loop enclosing an area A lies in a uniform magnetic field B.  The magnetic flux through the loop is equal to BA*cos(theta),  hence: induced emf can further be expressed as:

 (*note, we need to multiply this by N turns)

 

Furthermore, Area of a clyinder wire is pi*r^2 and so we can substitute this into the formula and get:

Emf = -N*B*pi*r^2* d(cos(theta)/dt)

d(cos(theta)) is simply the change in cos, which is  

((cos(theta)final) - cos(theta)initial)) 

and dt is simply change in time, t:

therefore

Emf = -N*B*pi*r^2* ((cos(theta)final)-cos(theta)initial))  / t)

where, N = 32

r = (1.1 / 2 ) m

B =  47.0 µT = 47*10^-6 T

theta final = 180, theta initial = 0

t = 0.2 s

plugging this back into the formula for Emf we get:

Emf = - 32* (47*10^-6 T) * pi * ((1.1 / 2 ) m)^2* ((cos(180)-cos(0))  / (0.2 s))

Emf = 0.0143 V  =   14.3 mV

answered Feb 28, 2012 by pokemonmaster ~Expert~ (3,856 points)
selected Feb 28, 2012 by Joey33

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