This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Massachusetts Institute of Technology — Physics Department Physics 8.02 March 15, 2002. Non—Conservative Fields — Do Not Trust Your Intuition! The following notes may help you to digest the very nonintuitive consequences of Faraday’s Law as
discussed and demonstrated in my lecture of Friday, March 15. changing magnetic ﬂux A magnetic ﬁeld is present in the shaded area; it is perpendicular to the page, and it is changing in time.
There are two identical voltmeters V1 and V2, and two resistors R1 and R2; the internal resistance of each
voltmeter, R1, is much much larger than R1 and R2. All connecting wires have a negligible resistance. We identify three closed loops in this circuit: the left loop with voltmeter V1 and resistor R1, the middle
loop with the two resistors, and the right loop with voltmeter V2 and resistor R2. We call the currents in
these loops I1 (t), I (t), and I2 (t), respectively (see the diagram). We will assume that at time t the current
in each loop is clockwise. If one (or more) of our currents turns out to be negative, it simply means that
that current is counter clockwise. In both the left and right closed loop we may apply Kirchhoff ’s 2nd rule: fﬁﬁzo (1) However, in the middle closed loop this rule can NOT be used; instead we now MUST use Faraday’s Law
which ALWAYS holds, we could also have used Faraday’s Law for the left and right loops (the right side of
Faraday’s Law would then have been zero): fﬁ.d7=5(t)=—“ﬁ (2) NOTICE: the induced EMF, 5, only depends on the magnetic ﬂux change through any open
surface that we attach to the closed loop, thus it depends on the change in the B ﬁeld in the
shaded area (since we may choose any open surface, I suggest we choose a ﬂat surface that
lies in the plane of our paper). 5 is therefore independent of R1 and R2. Applying the above equations, starting at A, and going clockwise around each loop separately, we have: Left loop (eq. 1) LR, + 11R1 — IR1 = 0 (3) lOOp (eq. 2) IR1 + IR2  I1R1  I2R2 Z 5 Right loop (eq. 1) I2R2 — IR2 + 12R, 2 0 (5) Bi >> R1 and Bi >> R2, therefore I1 << I and I2 << I, and these equations can be simpliﬁed: 11R, — IR1 m 0 (6)
I(R1+R2)~5 (7)
IgRi  IR2 N 0 Each voltmeter will show a reading which depends on the current through that meter (I1 through the
voltmeter on the left, and I2 through the voltmeter on the right). The scales of the meters have been
calibrated to show a voltage which is the product of the current through the meter and the internal
resistance of that meter. Thus the left voltmeter will read V1 2 LR, , and the right voltmeter will read
V2 = I2Ri. If we substitute this in eqs. (6) and (8) we ﬁnd: V1 Z IlRi N IR1 and
V2 Z IgRi N 1R2 If we connect the “+” side of both voltmeters with the A—side in the diagram (and the “—” side with the
D—side), the recorded values of V1 and V2 at any point in time will be opposite in sign, as demonstrated in
lectures. This is immediately obvious when you realize that a clockwise current I dictates a clockwise
current I1 as well as a clockwise current I2 (this follows from eqs. 6 and 8; I, I1 and I2 always have the
same sign). This means that when the current goes through the left voltmeter from its “+” side (A) to its
“—” side (D), thus when it will indicate a “positive” voltage, that the current through the right voltmeter
then goes from its “—” side (D) to its “+” side (A), and thus its voltage reading will be “negative”. If R1 and R2 are known, for given 5 (at a given point in time; see equation 2), the current I can be
calculated from eq. (7), and the voltage readings follow from eqs. (9) and (10), but they will have an
opposite sign. Notice that V2 / V1 N R2 / R1 (independent of I Escample Suppose: E = 1 Volt (at a given instant in time when the magnetic ﬂux, coming out of the paper, is
increasing), R1 = 100 9, R2 = 900 Q, and R, = 107 Q, then, using eq. (7), we ﬁnd I m 1.0 x 10—3 A
(clockwise), and using eqs. (9) and (10), we ﬁnd V1 x 11—0 Volt, and V2 x 19—0 Volt, and the polarities of
V1 and V2 are opposite! In case you are interested in I1 and I2, using eq. (9), we ﬁnd I1 m 1.0 x 10—8 A, and using eq. (10), we ﬁnd
I2 m 9.0 x 10—8 A. Notice how small they both are in comparison with I. Both I1 and I2 are in clockwise
direction. If R1 were 5 Q, and R2 were 45 9, (thus both 20 times smaller than above), I would be about 2.0 x 10—2 A
(20 times higher than above), but the values for V1 and V2 would be the same as before! Thus, for the given ratio R2 /R1 = 9, at any time  V2 x 19—05 and  V1 m %5, but the polarities are always opposite! Summary. If I “travel” from A to D through the resistor R1, the “potential difference”1 between A and D is m IR1 (A
has a higher “potential”1 than D), and this value (V A — VD) is registered by the left voltmeter. If I
continue my journey through R2, back to A, the “potential difference”1 between D and A is m IR2 (D has
a higher “potential”1 than A), and this value (VD — V A) is registered by the right voltmeter. Thus,
once I have completed the closed loop journey in the middle loop, starting at A, and ending at A, the
“potential difference”1 V A — V A 75 0. Isn’t that weird? NO, notice eq. (2)! In NOTL Conservative Fields, the electric potential diﬁerencel, if one deﬁnes this as the integral of?)  d l between two p0ints, depends on the path, our intuition breaks
down completely. Test yourself. Test 1. If you want to see whether you understand this difﬁcult concept, calculate what the relative
readings of V1 and V2 are for the diagram below. Notice that the wire that was connecting the left
voltmeter with the D—side in our diagram on page 1, now is again connected with the D—side, but it is
wrapped once around the whole circuit. The small arc in the wire above R1 indicates that it is not in
contact with the horizontal wire with which it appears to intersect due to the 2—dimensional projection. To
stress this further I have interrupted the horizontal wire a triﬂe on either side of the arc. This interruption
is not real; the horizontal wire is continuous. changing magnetic ﬂux Test 2. Now wrap that same wire not once around the whole circuit but 100 times before you connect it
again on the D—side. Without realizing it, in doing so, you have been building yourself some kind of a
transformer (we will discuss transformers later in the course). What now will the relative readings be
between V1 and V2? I hope this was helpful, this is not easy! Walter Lewin 1In nonconservative ﬁelds it may be better not to use the words “potential diﬁerence,” but instead, one should say “the
. . —> ~ »
line integral of 3 ~ dl along a speCiﬁed path. ...
View
Full
Document
This note was uploaded on 04/23/2011 for the course PHYS 1441 taught by Professor White during the Spring '08 term at UT Arlington.
 Spring '08
 WHITE
 Physics

Click to edit the document details