Hi:
I am trying to understand the reasons behind the signs in Faraday's Law.
Say there is a single loop of wire (N=1) on a horizontal surface (like
the paper) through which a magnetic flux density B is increasing in the
upward direction. The loop is almost closed, but a resistor load is
connected across the small gap in the loop.
We will define the surface normal ds to be up, so that the contour C of
the loop, is counterclockwise. We also define V_tr_emf (the induced
voltage across the gap in the stationary loop by a changing magnetic
field) to be positive where when following the contour we first
encounter the gap, and negative when we encounter the other side of the gap.
Since B is increasing up and given the definitions established, it is
apparent that assuming a uniform B always perpendicular to the surface,
the quantity N*d(flux)/dt is negative, and so the real voltage measured
by a voltmeter at the gap would be opposite the definition of V_tr_emf, so:
V_tr_emf = -N*d(flux)/dt , which is the definition of Faraday's Law as
given in my EM text. Things are looking good so far.
+-------------------+
+--------+ | |
| | | loop: |
| | | B increasing |
/ (-) +-------+ 1 out of paper |
\ (+) toward you, | i current points
/ R V_tr_emf | | down in accordance
\ (-) | | with Lenz' Law
/ (+) +-------+ 2 ds points | V
| | | out of paper |
| | | toward you |
+--------+ | |
+-------------------+
C---> contour is CCW
Now we know that the E field in the resistor is with the direction of
current flow. But the current flow in the loop is against the direction
of V_tr_emf. Remember that the *real physically observed* V_tr_emf has
the opposite sign of the *defined* V_tr_emf.
What is the direction of E in the loop (E_loop)?
I would think that E_loop must be in agreement with the real physical
value of V_tr_emf, which would make E point opposite the direction of
current flow in the loop. This is also consistent with the fact that
the loop is a generator, so the magnetic field is importing energy to
the charge carriers, thus, the current must be moving *against* E_loop
in the loop, while in the external circuit current is moving *with* E
through the resistor, where it is dissipating energy.
But if that is so then the contour integral around C from point 2 to
point 1 of E_loop is positive.
Thus we have:
integral_C(E_loop . dl_vec) > 0
and
N*d(flux)/dt > 0
Thus we have the opposite sign on one or the other terms compared to
what is known in Faraday's Law (the full relation in MAxwell's equations:)
integral_C(E_loop . dl_vec) = - N*d(flux)/dt
On the other hand, if we say that the E_loop field is in accordance with
the defined direction of V_tr_emf, then the quantity
integral_C(E_loop . dl_vec) < 0
which agrees with what we know about the other side of Faraday's Law.
So the question is which way does E_loop point?
I really expect it to point opposite the direction of i_loop, as charge
must be gaining potential as it traverses the loop. This is also
consistent with Kirchoff's Law, or:
integral_C_closed(E . dl_vec) = 0 around the entire circuit, which can
only be true if E_loop is against the current in the loop, while E is
with the current in the resistor.
But with this interpretation, the math of Faraday's Law has the wrong sign.
What is wrong with this picture?
Thanks.
--
_______________________________________________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
crcarle@sandia.gov -- NOTE: Remove "BOGUS" from email address to reply.
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