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A magnetic field of 0.55 g is directed straight down, perpendicular to the plane of a circular coil of wire that is made up of 550 turns and has a radius of 20 cm. 1) if the coil is stretched, in a time of 35 ms, to a radius of 50 cm, calculate the emf induced in the coil during the process.

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skyluke89
The emf induced in the coil is given by Faraday-Neumann-Lenz law:
[tex]\epsilon = - \frac{\Delta \Phi}{\Delta t} [/tex]
where 
[tex]\Delta \Phi[/tex] is the variation of magnetic flux through the coil
[tex]\Delta t[/tex] is the time interval

The magnetic field intensity is
[tex]B=0.55 G \cdot (1 \cdot 10^{-4} T/G) = 0.55 \cdot 10^{-4} T[/tex]

Since the magnetic field strength is constant, the variation of flux through the coil is given by
[tex]\Delta \Phi = N B \Delta A[/tex] (1)
where N=550 is the number of turns, while [tex]\Delta A[/tex] is the variation of area of the coil. We can re-write (1) as 
[tex]\Delta \Phi = NB (\pi r_1^2 - \pi r_2^2)=(550)(0.55 \cdot 10^{-4} T)(\pi (0.20m)^2-\pi (0.50m)^2) =[/tex]
[tex]=-0.020 Wb[/tex]

The time interval is [tex]\Delta t=35 ms=0.035 s[/tex], therefore the induced emf is
[tex]\epsilon = - \frac{\delta \Phi}{\Delta t}=- \frac{-0.020 Wb}{0.035 s}=0.57 V [/tex]

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