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Unformatted text preview: Physics , 8.02 Massachusetts Institute of Technology April 1, 2002 Almost all authors of college physics books are confused about the proper use of Faraday’s Law in circuits
with inductors. Giancoli is no exception. Professor John Belcher (who has lectured 8.02 many times) has
a wonderful Lecture Supplement which sets the record straight. It follows below. I (Walter Lewin) have
amended it slightly by adding references to Giancoli (Belcher used a different book which made the same
embarrassing mistakes) and by referencing my 8.02 lecture of March 15, 2002. It may help to ﬁrst read the
lecture supplement on non—conservative ﬁelds of March 15. SelfInductance — Kirchhoff ’s 2nd Law — Faraday’s Law
The addition of time—changing magnetic ﬁelds to simple circuits means that the closed line integral of the
electric ﬁeld around a circuit is no longer zero. Instead, we have, for any open surface feasZJ/ﬁdn
dt Any circuit where the current changes with time will have time—changing magnetic ﬁelds, and therefore
induced electric ﬁelds. How do we solve simple circuits taking such effects into account? We discuss here
a consistent way to understand the consequences of introducing time—changing magnetic ﬁelds into circuit
theory — that is, inductance. As soon as we introduce time—changing magnetic ﬁelds, the electric potential difference between two
points in our circuit is no longer well—deﬁned, because when the line integral of the electric ﬁeld around a
closed loop is no longer zero, the potential difference between points a and b, say, is no longer independent
of the path used to get from point a to point b. That is, the electric ﬁeld is no longer a conservative ﬁeld,
and the electric potential is no longer an appropriate concept (fl can no longer be written as the negative
gradient of a scalar potential). However, we can still write down in a straight—forward fashion the equation
that determines the behavior of a circuit. To show how to do this, consider
the circuit shown in the sketch to the
right. We have a battery, a resistor, a
switch S that is closed at t = 0, and
a “one—loop inductor.” It will become
clear what the consequences of this “in—
ductance” are as we proceed. For t > 0,
current will ﬂow in the direction shown
(from the positive terminal of the bat—
tery to the negative, as usual). What is
the equation that governs the behavior
of our current i for t > 0? +  Switch S _
closed at t: 0 l To investigate this, apply Faraday’s Law to the open surface bounded by our circuit, where we take d_A out of the page, and (Ts right—handed
with respect to that choice (counter—clockwise). First, what is the integral of the electric ﬁeld around this
circuit? Well, there is an electric ﬁeld in the battery, directed from the positive terminal to the negative
terminal, and when we go through the battery in the direction of (is that we have chosen, we are moving
against that electric ﬁeld, so that E . (Ts is negative. Thus the contribution of the battery to our integral is
— 5. Then, there is an electric ﬁeld in the resistor, in the direction of the current, so when we move through
the resistor in that direction, fl . (Ts is positive, and that contribution to our integral is + iR. What about
when we move through our “one—loop inductor”? There is no electric ﬁeld in this loop if the resistance of
the wire making up the loop is zero (this may bother you — if so, see the next section). If the wire has a
small resistance 7" << R, then there will be an electric ﬁeld in the wire, and a contribution to the integral of
the electric ﬁeld of +17", which we just lump with the 1R term we already have (that is, we redeﬁne R to
include both resistances). So, going totally around the closed loop, we have fﬁszﬁ+m Now, what is the magnetic ﬂux (b through our open surface? First of all, we arrange the geometry so
that the part of the circuit which includes the battery, the switch, and the resistor makes only a small
contribution to (b as compared to the (much larger in area) part of the open surface which includes our “”one—loop inductor Second, we know that (b 1s positive in that part of the surface, because current ﬂowing
counter—clockwise will produce a B ﬁeld out of the paper, which 1s the same direction we have assumed for
dA, so B dA 1s positive Note that B 1s the self magnetic ﬁeld — that 1s the magnetic ﬁeld produced by
the current ﬂowing 1n the circuit, and not by any external currents. —; We also know that at any point in space, B is proportional to the current i, since it is computed from
the Biot—Savart Law, to wit, ‘ ,u_0ds><r
B:
Z 4—71" r3 If we look at this expression, although for a general point in space it involves a very complicated integral
over the circuit, it is clear that B is everywhere proportional to i. That is, if we double the current, B at
every point in space will also double, all other things being the same. It then follows that the magnetic ﬂux
(b itself must also be proportional to i, since it is the surface integral of B, and B is everywhere proportional
to i. That is, we must have (b = Li, where L is a constant for a given arrangement of the wires of the circuit.
If we change the geometry of the circuit (i.e., suppose we halve the radius of the circle in our sketch), we
will change L, but for a given geometry, L does not change. The quantity L is called the self—inductance of
the circuit, or simply the inductance. From its deﬁnition, we can show that the inductance has dimensions
of no times length. We give an estimate of L for a single loop of wire below. But ﬁrst, let us write down the equation that governs the time evolution of i. If (b = Li, then the time
rate of change of (b is just L di/dt, so that we have from Faraday’s Law
.. _. , d¢ di fE.ds_ €+ZR— dt_ Ldt (1) If we divide equation (1) by L, and rearrange terms, we ﬁnd that the equation that determines the behavior of i is di/dt+(R/L)i = 5/ L. The solution to this equation given our initial conditions is i(t) = (s/R)(1 —e_Rt/L) [see Giancoli, equation 30—9, p. 762]. This solution for i(t) reduces to what we expect as t gets very large, s/R, but also shows a continuous rise of the current from zero at t = 0 to this ﬁnal value, in a characteristic time TL 2 L / R (TL is called the inductive time constant). This is the effect of having a non—zero inductance in a circuit, i.e., of taking into account the induced electric ﬁelds due to time changing B ﬁelds. And this is what we expect from Lenz’s Law — the reaction of the system is to try to keep things the same, that is to delay the build—up of current (or its decay, if we already have a current ﬂowing in the circuit). Kirchhoff ’5 Second “Law” Modiﬁed for Inductors We can write the governing equation for i(t) from above (equation (1)) as di
+5—iR— Ld—zAV20 (2)
where we have now cast it in a form that “looks like” a version of Kirchhoff ’s Second Law, namely that the
sum of the potential drops around a circuit is zero (we are still moving counter—clockwise around the circuit,
but the overall sign changes from equation (1) to (2) because we are now adding up changes in electric
“potential” ). Our text (Giancoli) chooses to approach circuits with inductance by preserving “Kirchhoff ’s Second Law,”
or the loop theorem, by specifying the “potential drop” across an inductor. To get the correct equation,
Giancoli must make an additional “rule” for inductors as follows: Inductors: If an inductor is traversed in the direction moving with the current, the change
in potential is —L di/dt; if it is traversed in the direction opposite the current, the change in
potential is +L di/dt. Although Giancoli never explicitly states this rule, it is implicit in his use of the “Loop Theorem” in sections
30—4, 30—5, and 30—6. Use of this formalism will give the correct equations. However, the continued use of Kirchhoff ’s Second
Law with this additional rule is MISLEADING at best, and at some level DEAD WRONG in terms
of the physics, for the following reasons. Kirchhoff ’s Second Law was originally based on the fact that the
integral of E around a closed loop was zero. With time—changing magnetic ﬁelds, this is no longer so, and
thus the sum of the “potential drops” around the circuit, if we take that to mean the negative of the closed
loop integral of E, is no longer zero — in fact it is +L di/dt. As do many introductory texts, Giancoli
brings the Ldi/dt term to the other side of the equation, adds it to the negative of the closed loop of E, and
ascribes it to a “potential drop” across the inductor. This approach gives the right equations, but it sure confuses the physics. In particular, having a “potential
drop” across the inductor of —L di/dt implies that there is an electric ﬁeld in the inductor such that the integral of E through the inductor is equal in magnitude to Ldi/dt. This is not always, or even usually, true, as in our example above (the integral of E through our “one—loop” inductor above is zero, NOT
L di/dt). The fact that E is zero in our “one—loop inductor” above may confuse you, and for good reason. You have
developed some intuition about induced electric ﬁelds, based on the kinds of Faraday’s Law problems we
have been doing up to now. The fact is, quite often in the past when we have had time—changing magnetic
ﬁelds, we have had an electric ﬁeld right where dE/dt was non—zero. That fact would make you think that
Giancoli is right, that there is an electric ﬁeld right there in the inductor, and thus a potential drop across
it. What has changed in our circuit above to make E zero in our “one—loop inductor,” even though there is
a time—changing magnetic ﬁeld through it? This is a very subtle point, and the source of endless confusion,
so let’s look at it carefully. O ' t 't' th 13
thereulshlodldloble a: G ﬂ G G electric ﬁeld in an in— ductor is based on do— ing problems like that G G G
shown in the sketch a
to the right. We have a loop of wire of ra— G G G G
dius a and total resis—
tance R, immersed in
an external magnetic G G G G 2 4 ﬁeld which is out of I (sec) the page and increasing with time as shown. In considering this circuit, unlike our “one—loop” inductor
above, we neglect the magnetic ﬁeld due to the currents in the wire itself, assuming that Eext is much greater
than that ﬁeld, and consider only the effects of the external ﬁeld. The conclusions we arrive at here can be
applied as well to the self—inductance case. The changing external magnetic ﬁeld will give rise to an induced electric ﬁeld in the loop of wire, with
a line integral which is equal to —7m2 (dBext /dt). This induced electric ﬁeld is azimuthal and uniformly
distributed around the loop (see sketch). We have from Faraday’s Law that —; ~ dB t
2Trailllinduced : _7Ta2 d:x ‘ adﬁext
=>E~ = —— induced 2 dt Thus if the resistance is distributed uniformly around our wire loop, we get a uniform Einduced in the
loop which is the same at every point in the wire loop, and circulating clockwise for Bext increasing in time. This electric ﬁeld causes a current, with the cur— \\\\\\\\\\ \ E rent density given by j: Einduced/r (the microscopic form of Ohm’s Law). The total current in the wire loop will
be the total “potential drop” around the loop divided by
the resistance R (the macroscopic form of Ohm’s Law),
or 27raflinduced/R. This current will circulate clockwise in the same sense as Einduced. Thus if the resistance is dis— tributed uniformly around our wire loop, we get a uniform
Einduced in the loop which is the same at every point in the wire loop, and circulating clockwise for increasing Bext. But what happens to the electric ﬁeld if we don’t dis—
tribute the resistance uniformly around the loop. For ex—
ample, let’s make the left half of our loop out of wire with
resistance R1, and the right half the loop out of wire with
resistance R2, with R = R1 + R2, so that we have the same total resistance as before (see sketch next page).
Let’s furthermore assume that R1 < R2. NOTICE some similarity with the demo I (Walter Lewin) did in
lecture on March 15, 2002 (read my Lecture Supplement). How is the electric ﬁeld distributed around the
loop of wire now? First of all, the emf in the circuit is the same as above, as is the total resistance, so that
the current i has to be the same as above. Moreover it must be the same on both sides of the loop, by
7 charge conservation. But the electric ﬁeld in the left half
of the wire loop (E1) must now be diﬂerent from that in the right half (E2). Einduced This is so because the line integral of the electric ﬁeld
on the left side, over the left side, is 7raE1, and this must
be (from Ohm’s Law) equal to iRl. Similarly, 7raE2 = iR2.
Thus E1 /E2 = R1/R2, and therefore E1 < E2, since R1 <
R2. And this makes sense. We must get the same current
on both sides, even though the resistances are different. We
do this by adjusting the electric ﬁeld on the side with the
smaller resistance to be smaller. Because the resistance
is also smaller, we produce the same current as on the
opposing side with this smaller electric ﬁeld. But what happeneo to our uniform electric ﬁeld? Well, there are two ways to produce electric ﬁelds —
one_‘from time—changing magnetic ﬁelds, the other from electric charges. Nature accomplishes the reduction
in E1 compared to E2 by charging up the junctions separating the wire segments (see sketch above), positive
on top and negative on bottom. _‘The total electric ﬁeld is the sum of the electric ﬁeld induced by the
changing external magnetic ﬁeld (Einduced, as indicated in the sketch above, still clockwise), and the electric
ﬁeld associated with the charging at the junctions (Echarge, as indicated in the sketch, going from positive
charge to negative charge, as is always true for ﬁelds produced by charges). It is clear that the addition of
these two contributions to the electric ﬁeld will reduce the total electric ﬁeld on the left and enhance it on
the right. The ﬁeld E1 will always be clockwise (as it must be to produce clockwise current ﬂow), but it
can be made arbitrarily small by making R1 << R2. However, we still always have the integral of f} over the
complete closed loop equal to —7m2 (dBext / dt), as Faraday’s Law demands. Thus we see that we can make a non—uniform electric ﬁeld in an inductor by using non—uniform resistances,
even though our intuition tells us (correctly) that the induced electric ﬁeld should be uniform at a given radius.
The reality is that there is another way to produce electric ﬁelds, namely from charges, and Nature uses that
fact as needed. All that Faraday’s Law tells us is that the line integral of E around a closed loop is equal
to the negative time rate of change of the magnetic ﬂux through the enclosed surface. It doesn’t tell us at what locations the E ﬁeld is non—zero around the loop, and it may be non—zero (or zero!) in unexpected
places. The ﬁeld in wire making up the “one—loop inductor” above was zero (or at least very small), with
the signiﬁcant ﬁelds occurring only in the resistor and the battery, for exactly the sort of reason we have
considered here. One ﬁnal point. Suppose you put the probes of a voltmeter across the terminals of an inductor (with very
small resistance) in a circuit. What will you measure? What you will measure on the meter of the voltmeter
is a “voltage drop” of Ldi/dt. But that is not because there is an electric ﬁeld in the inductor! It is because
putting the voltmeter in the circuit will result in a time changing magnetic ﬂux through the voltmeter
circuit, consisting of the inductor, the voltmeter leads, and the large internal resistor in the voltmeter (see
my Lecture Supplement of March 15, 2002). A current will ﬂow in the voltmeter circuit because there will
be an electric ﬁeld in the large internal resistance of the voltmeter, with a potential drop across that resistor
of L di/dt, by Faraday’s Law applied to the voltmeter circuit, and that is what the voltmeter will read. The
voltmeter as usual gives you a measure of the potential drop across its own internal resistance, but this is not
a measure of the potential drop across the inductor. It is a measure of the time rate of change of magnetic
ﬂux in the voltmeter circuit! As before, there is only a very small electric ﬁeld in the inductor if it has a
very small resistance compared to other resistances in the circuit. If you ﬁnd all this confusing, you are in good company. This is one of the most difﬁcult
and subtle topics in this course — it trips up experts all the time. Not easy! ...
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 Physics

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