Amplitude of the induced voltage across the coil

Thread Starter


Joined Mar 30, 2010
If anyone can advise that would be great.

1. The plane of a 5 turn coil of 5mm² cross sectional area is rotating a 1200 r.p.m in a magnetic field of 10mT.

Q. Calculate the amplitude of the induced voltage across the coil


5 turns
c = 1200 rpm
A = 5cm²
B = 10 mT

2. φ = N.B.A φ = 5×10×10^-3 ×5 × 10^-2
φ = 2.5 ×10^-3

V = dφ/dt or N.A. dB/dt

The equation thought to have been used is the following but the Length is not vissible to me.

V/L = c x B c being the speed
B field
L lenth of wire

3. I have tried two avenues please advise

a: V(t) = -N•d(B•A)/dt = -N•B•dA/dt = N•B•ω•A•sin(ωt)
V(t) =(5)(10x10^-3)(125.66370599999999)(5)sin(ωt) = 31.415 sin(ωt) volts.


b: And using the following, saying t = 60 sec

Then φ = N.B.A φ = 5×10×10^-3 ×5 × 10^-2
φ = 2.5 ×10^-3

And V = dφ/dt or N.A. dB/dt = 4.17x10^-3 V

My concern here is that 1200rpm was not used.


Joined Mar 6, 2009
For simplicity I would assume a rectangular cross section for the coil.

Suppose the rectangular side length orthogonally cutting the field is L and the top & bottom length is D. The top & bottom sides are assumed not to be cutting the field and produce no emf.

The emf producing side of length L sweeps through a circular path length of ∏*D at 1200 rpm or 20 rotations per second. The side L then cuts the field at a maximum velocity of

v=∏*D*20 m/sec

The emf per side is then of the form


AND the total emf = 2*e

You know the relationship between L, D and the given coil cross sectional area [CSA].

You should be able to find Emax & ω from the derived rotational velocity v and the other information supplied. Emax will be in terms of L & D which you would then relate to the coil CSA.

1. The general relationship for induced emf is E=B*L*v for a conductor of length L cutting a field of B Tesla at constant velocity v meters/sec.

2. In this case the output is a sinusoid since either side of length L is cutting the field at a sinusoidally varying rate throughout the total cycle. But you do know the maximum rate (or velocity) at which the field is being cut.

NB - There are 5 turns per side - not just one!
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