- #1
Zynwyx
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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?
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?