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jearls74

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- Thread starter jearls74
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- #1

jearls74

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- #2

jearls74

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I believe i have figured out the answer to my question. THE B-field is contained in the stator core and since the permeability of air is very poor ( like 1 i believe) the field only affects the wire that passes through the stator core, so its not total length of the wire in each phase, but only the wire that passes through the core. If this is correct, it leads me to a new question about the length of the wire: would i multiplythe length of wire that passes through the stator times the number of wires in each coil? could someone tell me if im right or wrong please.

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- #3

Bob S

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Bob S

- #4

jearls74

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Hi everyone, ive been looking into faradays law of induction on the advice of Bob and a question about the equation. In the attached picture, the AC generator example, how is the area calculated? Im not exactly following the meters squared per second part? i looked up the definition on wikipedia and understand that its speed or velocity defined by distance in meters per second. Is the Area the total distance traveled by the rotor in one second? any help is appreciated.

- #5

Bob S

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Use

N=100

B=0.4 Tesla (static field)

A = 5 cm by 5 cm (=0.05m x 0.05m = 0.0025 m

RPM = 600; rps = 10 Hz; ω = 62.8 radians per second.

So V(t) = -N·d(B·A)/dt = -N·B·dA/dt = N·B·ω·A·sin(ωt)

=(100)(0.4)(62.8)(0.0025)sin(ωt) = 6.28 sin(ωt) volts.

Bob S

- #6

lee.perrin@gm

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Use

N=100

B=0.4 Tesla (static field)

A = 5 cm by 5 cm (=0.05m x 0.05m = 0.0025 m^{2}) rotating coil

RPM = 600; rps = 10 Hz; ω = 62.8 radians per second.

So V(t) = -N·d(B·A)/dt = -N·B·dA/dt = N·B·ω·A·sin(ωt)

=(100)(0.4)(62.8)(0.0025)sin(ωt) = 6.28 sin(ωt) volts.

Bob S

Hi I read through your discussion and seem to think that this might be slightly similar to my question. ( rotating coil in a magnetic field )

And the equation that I have been trying to use limits me as I do not know the length.

The picture on the bottom right mentioned. Is Vgenerated the same as Vinduced?

V/L = c x B

c being the speed

B field

L lenth of wire

Here is the question:

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.

Any info whould be great.

Lee

- #7

lee.perrin@gm

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So 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

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.

- #8

jearls74

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- #9

jearls74

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Well, now that i understand how to calculate the induced voltages in the stator, im having a problem understanding how to position the magnets correctly. I am going to use 12 magnets (6 pole pairs), by dividing 360 degrees by 12 magnets i get 30 degrees for each magnet. Using the stator diagram as reference, 30 degrees for each magnet, the magnet covers one side of each coil in all three phases, is that the correct placement? any insight would be appreciated

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