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Science Forum Index » Physics - Electromagnetic Forum » Simple Faraday's Law question.
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| Cyberkatru |
 Posted: Fri Dec 22, 2006 2:46 pm |
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If there is a changing flux through a loop of conducting wire then a
current is produced. Lens' law says that this current produces a B
field that opposes the change. But wait, should this new B field be
included in with the original whose changing flux is causing the
current in the first place? (nonlinear feeback) If so then plenty of
elemetary solutions found in physics text seem wrong.
For example, the usual solution to the problem of a rotating loop in a
uniform B field doesn't seem to take this induced opposing field into
account. |
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| Edgar A Pearlstein |
| Posted: Fri Dec 22, 2006 2:46 pm |
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Cyberkatru (cyberkatru@gmail.com) wrote:
: If there is a changing flux through a loop of conducting wire then a
: current is produced. Lens' law says that this current produces a B
: field that opposes the change. But wait, should this new B field be
: included in with the original whose changing flux is causing the
: current in the first place? (nonlinear feeback) If so then plenty of
: elemetary solutions found in physics text seem wrong.
Yes. The effect is called self-inductance, and must be taken into
account for highest accuracy.
:
: For example, the usual solution to the problem of a rotating loop in a
: uniform B field doesn't seem to take this induced opposing field into
: account.
In a DC motor, this is called "back emf". Such a motor takes more
current when starting (i.e. not yet up to speed) than when going at full
speed.
Older physics textbooks discuss these things, but newer ones tend not
to. |
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| Wimpie |
| Posted: Sat Dec 23, 2006 4:31 pm |
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Cyberkatru ha escrito:
Quote: If there is a changing flux through a loop of conducting wire then a
current is produced. Lens' law says that this current produces a B
field that opposes the change. But wait, should this new B field be
included in with the original whose changing flux is causing the
current in the first place? (nonlinear feeback) If so then plenty of
elemetary solutions found in physics text seem wrong.
For example, the usual solution to the problem of a rotating loop in a
uniform B field doesn't seem to take this induced opposing field into
account.
Hello,
Yes, you are right. The voltages induced in your short-circuited loop
causes a current. That current generated a new flux. Part of that flux
goes through the driving coil and interferes with the flux through the
driven loop. So there is a linear feedback mechanism.
Consider two loops A and B with negligible resistance. The loops are
spaced about 25% of the diameter. Loop A is driven with a constant AC
current. The voltage across loop A is measured. The voltage across
the open loop B will be lower, because not all of the flux from loop A
passes through loop B.
Now short circuit loop B. The voltage across loop B reduces to zero,
and therefore the net flux does also. You are correct, the flux through
loop B, generated by the current through loop B, cancels the original
flux in loop B.
Now some part of the flux generated by loop B passes through loop A,
reducing the net flux in loop A. This reduction in flux results in a
reduction of the voltage across loop A.
When you should feed loop A with a constant voltage source, the flux
through loop A must be constant (because of EMF = d(flux)/dt). In that
case, short circuiting loop B will result in an increase in the current
through loop A (maintaining the flux through loop A constant). This
happens in a transformer. As you load the secondary coil, the current
in the primary will increase.
In general when you do not change the frequency, the AC flux through a
loop is solely determined by the AC voltage across the perfectly
conducting loop, no matter whether the AC voltage is caused by a source
or the AC voltage is induced by another changing B-field.
Of course in real world there is also wire resistance. The voltage
across the (distributed) wire resistance must be added to the induced
voltage. Depending on the direction of current, the voltage across the
wire resistance tends to increase or decrease the actual voltage across
the loop.
I hoop this will help you.
Best regards,
Wim
PA3DJS. |
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