Induced EMF in a spinning wheel

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Main Question or Discussion Point

Faraday's law says that an induced voltage across a conductor is directly proportional to the rate of change of magnetic flux through the conductor, but I am having trouble getting my head round how it works in this situation.

Imagine a perfectly circular thin copper disc, and that this disc is perpendicular to a B-field, i.e. the lines of magnetic flux run through the face of the disc at a 90 degree angle to the plane of the disc.

|
|------<------- (magnetic flux lines are the dotted lines, the | is the disc viewed side-on)
|------<-------
|------<-------
|------<-------
|------<-------
|

The copper disc is rotated at a constant speed around its centre; and thus a voltage is induced across it, from its outer edges to its centre. But how is this voltage induced? I can understand that initially when the disc is spun to accelerate it up to the constant speed at which it will be spun, d[tex]\phi[/tex]/dt increases and so a voltage is induced. But once the disc is spinning at a constant speed, if the disc is perfectly circular, then it is radially symmetrical; and so no matter at what angle the disc is rotated through, as long as the disc is being rotated at this constant speed shouldn't the magnetic flux through each portion of the disc be constant? From the reference point of a point on the face of the disc, wouldn't a constant amount of flux flow through it when it is moving at a constant speed, just like how a constant amount of flux flows through it when the disc is stationary?

I think my flaw in thinking here is that I think that at a distance r from the centre of the disc, the B-field is constant. But I cannot see how it can be any other way. So as a point on the disc travels out a circular path, wherever it is on that circular path the flux through it is constant, and therefore the rate of change of flux is 0. So no e.m.f should be induced, but according to the exam question where I first saw this problem there is... hopefully someone could clarify where I've got this wrong?
 

Answers and Replies

  • #2
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I think there is a simpler "microscopic" way to think about this: Consider the detailed properties of the metal disk. It consists of positive and negative charges.

These charges are moving in a B-field. Thus the Lorentz force acts on those moving charges. Depending on the direction of rotation, the positive charges might be pushed outwards while the nagative charges inwards (you know the right-hand rule for the Lorentz force, right?).

Thus there is a radial E-field and a radial potential difference.

Isn't this the reason you also need to consider the amount of magnetic flux that is "cut" per unit time to determine induced currents when applying Faraday's law?
 
  • #3
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Dr. Munley explained this in detail. The B field is the mag flux density. The quantity "phi" is the mag flux linkage. So, the induced emf is given by v = -N*d(phi)/dt.

Although B is fixed, phi is varying w/ time to the 1st power. The Lorentz force on the free electrons are center to rim in a radial direction, or the negative depending on polarity of B & rotation of disk.

The closed loop element area of the electron paths is a wedge shape. Since the galvanometer has one terminal connected at the center w/ the other connected at the rim, the electron starts at the center & moves outwards due to Lorentz force. This is happening throughout the disk. The closed loops are formed starting at the center, moving outward, but the disk has slightly rotated so that the electron reaches the outer rim "in front" of the contact on the rim. The arc length along the rim back to the meter contact completes the loop.

This loop area is equal to the area of the wedge shape sector. A complete circle is 2*pi radians, w/ an area of pi*r^2, "r" being radius. The angle of the wedge, theta, equals omega*t, where omega is angular speed, t is time. Thus phi = A*B = (theta/2*pi)*(pi*r^2)*B = (theta/2)*r^2*B. But theta = omega*t, so that phi = (omega*t/2)*r^2*B.

Rewriting, phi = ((omega*B*r^2)/2)*t. So phi varies w/ time to the 1st power. Since v = -N*d(phi)/dt, & N = 1 (one turn), then v = - (omega*B*r^2)/2. This suggests that the induced current & voltage are non-time-varying, i.e. "direct current". Observation comfirms this. The induced emf/current are measured to be dc.

I cannot find my copy of the Munley paper. I'll search, meanwhile if anyone has it, please send. Thanks. BR.

Claude
 

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