lecture 17 magnetic forcestorque, faraday’s law
TRANSCRIPT
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
1/33
EECS 117
Lecture 17: Magnetic Forces/Torque, Faradays Law
Prof. Niknejad
University of California, Berkeley
Universit of California Berkele EECS 117 Lecture 17 . 1/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
2/33
Memory Aid
The following table is a useful way to remember theequations in magnetics. We can draw a very goodanalogy between the fields a
E H J
D B V A
1 P M
aI personally dont like this choice since to me E and B are real and so the
equations should be arranged to magnify this analogy. Unfortunately the equations
are not organized this way (partly due to choice of units) so well stick with conven-
tion
Universit of California Berkele EECS 117 Lecture 17 . 2/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
3/33
Boundary Conditions for Mag Field
We have now established the following equations for astatic magnetic field
H = J B = 0
C
H d = S
J dS = I
SB dS = 0
And for linear materials, we find that H = 1B
Universit of California Berkele EECS 117 Lecture 17 . 3/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
4/33
Tangential H
The appropriate boundary conditions followimmediately from our previously established techniques
1
2
C
Take a small loop intersecting with the boundary andtake the limit as the loop becomes tiny
C
H d = (Ht1 Ht2)d = 0
Universit of California Berkele EECS 117 Lecture 17 . 4/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
5/33
Tangential H (cont)
So the tangential component of H is continuous
Ht1 = Ht2 11 Bt1 =
12 Bt2
Note that B is discontinuous because there is aneffective surface current due to the change inpermeability. Since B is real, it reflects this change
If, in addition, a surface current is flowing in betweenthe regions, then we need to include it in the abovecalculation
Universit of California Berkele EECS 117 Lecture 17 . 5/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
6/33
Normal B
1
2
Sn
Consider a pillbox cylinder enclosing the boundarybetween the layers
In the limit that the pillbox becomes small, we have
B dS = (B1n B2n)dS = 0
And thus the normal component of B is continuous
B1n = B2nUniversit of California Berkele EECS 117 Lecture 17 . 6/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
7/33
Boundary Conditions for a Conductor
If a material is a very good conductor, then well showthat it can only support current at the surface of theconductor.
In fact, for an ideal conductor, the current lies entirelyon the surface and its a true surface current
In such a case the current enclosed by even an
infinitesimal loop is finite
CH d = (Ht1 Ht2)d = Jsd Ht1 Ht2 = Js
This can be expressed compactly as
n (H1 H2) = JsBut for a perfect conductor, well see that H2 = 0, soH1t = Js
Universit of California Berkele EECS 117 Lecture 17 . 7/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
8/33
Hall Effect
J
B0
V0
When current is traveling through a conductor, at anyinstant it experiences a force given by the Lorentzequation
F = qE + qv
B
The force qE leads to conduction along the length of thebar (due to momentum relaxation) with average speedvd but the magnetic field causes a downward deflection
F = qxE0 qyvdB0Universit of California Berkele EECS 117 Lecture 17 . 8/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
9/33
Hall Effect: Vertical Internal Field
+ + + + + + +
+ + + + + + +
+
In steady-state, the movement of charge down (orelectrons up) creates an internal electric field which
must balance the downward pullThus we expect a Hall voltage to develop across thetop and bottom faces of the conducting bar
VH = Eyd = vdB0d
Universit of California Berkele EECS 117 Lecture 17 . 9/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
10/33
Hall Effect: Density of Carriers (I)
J
B0
VHJ
Since vd = Ex, and Jx = Ex, we can write vd = Jx/
VH =Jx
B0d
Recall that the conductivity of a material is given by = qN , where q is the unit charge
Since vd = Ex, and Jx = Ex, we can write vd = Jx/
VH =Jx
B0d
Universit of California Berkele EECS 117 Lecture 17 . 10/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
11/33
Hall Effect: Density of Carriers (II)
Recall that the conductivity of a material is given by = qN , where q is the unit charge, N is the density ofmobile charge carriers, and is the mobility of the
carriersVH =
JxB0d
qN
N = JxB0dqVH
= IB0dAqVH
Notice that all the quantities on the RHS are either
known or easily measured. Thus the density of carrierscan be measured indirectly through measuring the HallVoltage
Universit of California Berkele EECS 117 Lecture 17 . 11/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
12/33
Forces on Current Loops
F
F1 F2
B
B
x
y
za
b
I1
I2 F
Since the field is notuniform, the net forceis not zero. Note theforce on the sidescancel out
F1 = y I1I22a
d F2 = +yI1I2
2(a + b)d
F = F1 + F2 = y I1I2d2
1a
1b
Universit of California Berkele EECS 117 Lecture 17 . 12/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
13/33
Torques on Current Loops (I)
B0 B0
free rotation
about this axis
force down
force up
In a uniform field, thenet force on the cur-
rent loop is zero. Butthe net torque is notzero. Thus the loop
will tend to rotate.
T = r
F
F1 = I1B0dxF2 = +I1B0dx
F1 + F2 = 0
Universit of California Berkele EECS 117 Lecture 17 . 13/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
14/33
Torques on Current Loops (II)
F1
F2
I
IB0
x
yz
T1 =
(b/2)I1B0d sin z
T2 = (b/2)I1B0d sin z
T = T1 + T2 = zB0moment
I b d
Area of loopsin
In general the torque can be expressed as
T = m
B
where the moment is defined as m = I AreaUniversit of California Berkele EECS 117 Lecture 17 . 14/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
15/33
Electric Motors
B0A DC electric mo-tor operates on this
principle. A uniformstrong magnetic fieldcuts across a currentloop causing it to ro-tate.
When the loop is || to the field, the torque drops to zerobut the rotational inertia of the loop keeps it rotating.Simultaneously, the direction of the current is reversedas the loop flips around and cuts into the field. Thisgenerates a new torque that favors the continuousrotation.
Universit of California Berkele EECS 117 Lecture 17 . 15/
F d Bi Di
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
16/33
Faradays Big Discovery
In electrostatics we learned that
E d = 0Lets use the analogy between B and D (and E and H).Since q = Cv and = Li, and i = q = Cv, should we not
expect that = Li = v?
energy of field
converted to heat!
B
t
R
In fact, this is true!
Faraday was able toshow this experimentally
CEd =
d
dt = d
dt SBdSThe force is no longerconservative, E
=
Universit of California Berkele EECS 117 Lecture 17 . 16/
F d L i Diff ti l F
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
17/33
Faradays Law in Differential Form
Using Stokes Theorem
CE d = S
E
dS =
d
dt SB
dS =
SB
t dS
Since this is true for any arbitrary curve C, this impliesthat
E = Bt
Faradays law is true for any region of space, including
free space.In particular, if C is bounded by an actual loop of wire,then the flux cutting this loop will induce a voltage
around the loop.
Universit of California Berkele EECS 117 Lecture 17 . 17/
E l T f
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
18/33
Example: Transformers
V1
I1
V2
I2
+
+
L1 L2
MB
+
V1
+
V2
In a transformer, by definition the flux in the primaryside is given by 1 = L1I1
Likewise, the flux crossing the secondary is given by2 = M21I1 = M12I1 = M I1 (assuming I2 = 0)
Thus if the current in the primary changes, a voltage isinduced in the secondary
V2 = 2 = MI1
Universit of California Berkele EECS 117 Lecture 17 . 18/
G ti S k !
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
19/33
Generating Sparks!
++
V2
I1
I1
V2
t
t
large voltage!
large slope
Since the voltage at the secondary is proportional to therate of change of current in loop 1, we can generatevery large voltages at the secondary by interrupting thecurrent with a switch
Universit of California Berkele EECS 117 Lecture 17 . 19/
V t P t ti l
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
20/33
Vector Potential
Since B 0, we can write B = A. Thus
E =
B
t
=
A
t
=
A
tIf we group terms we have
E + At = 0So, as we saw in electrostatics, we can likewise write
E +A
t=
Universit of California Berkele EECS 117 Lecture 17 . 20/
M V t P t ti l
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
21/33
More on Vector Potential
We choose a negative sign for to be consistent with
electrostatics. Since if t = 0, this equation breaks
down to the electrostatic case and then we identify as
the scalar potential.
This gives us some insight into the electromagneticresponse as
E = At
E = electric response
At
magnetic response
In reality the EM fields are linked so this viewpoint is notentirely correct.
Universit of California Berkele EECS 117 Lecture 17 . 21/
Is the vector potential real?
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
22/33
Is the vector potential real?
We can now re-derive Faradays law as follows
V = CE d =
C
d
t CA
d
The line integral involving is zero by definition so wehave the induced emf equal to the line integral of A
around the loop in question
V = t
C
A d
We also found that equivalently
V =
tS
B dSUniversit of California Berkele EECS 117 Lecture 17 . 22/
The Reality of the Vector Potential
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
23/33
The Reality of the Vector Potential
V = t
C
A d
This equation is somewhat more satisfying thatFaradays law in terms of the flux. Although itsmathematically equivalent, it explicitly shows us the
shape of the loops role in determining the induced flux.
The flux equation, though, depends on a surfacebounding the loop, in fact any surface. Sometimes its
even difficult to imagine the shape of such a surface(e.g. a coil)
Universit of California Berkele EECS 117 Lecture 17 . 23/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
24/33
Vector Potential Outside of Solenoid
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
25/33
Vector Potential Outside of Solenoid
Then the voltage induced into the outer loop onlydepends on the constant flux generated within thecenter section coincident with the solenoid
Whats disturbing is that even though B = 0 along theloop, there is a force pushing electrons inside the outermetal.
The force is therefore not magnetic since B = 0.
The viewpoint with vector potential, though, does notpose any problems since A
= 0 outside of the loop.
Therefore when we integrate A outside of the loop,there is a nonzero result.
Universit of California Berkele EECS 117 Lecture 17 . 25/
Circuit Application of Transformers
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
26/33
Circuit Application of Transformers
V
+
2
+
V1
The transformer is avery important circuitelement
Before switching powersupplies, transformerswere ubiquitous in
voltage/currenttransformationapplications (taking wallvoltage of say 120V and
converting it to say 3V).
In fact, the name trans-former comes from this
very application
Universit of California Berkele EECS 117 Lecture 17 . 26/
Voltage Transformer
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
27/33
Voltage Transformer
V1 = L1dI1dt
+ MdI2dt
V2 = MdI1dt
+ L2 dI2dt
If I2
0, or for a light load on the secondary, we have
V1 = L1dI1dt
V2 = MdI1
dt
V2
V1 =
M
L1 = k
L1L2
L1 = kL2L1 = n
Universit of California Berkele EECS 117 Lecture 17 . 27/
Power Transmission
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
28/33
Power Transmission
Transformers are also used to boost the voltage for longrange power transmission
This follows since power loss proportional to I2R, so to
transmit a given power P, its best to use the largestvoltage to minimize the current I = P/V.
This is the reason we use AC power versus DC, since
transformers dont work with DC!
Universit of California Berkele EECS 117 Lecture 17 . 28/
Summary So Far
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
29/33
Summary So Far...
Lets summarize what weve learned thus far. There areno magnetic charges, so
B = 0and electric fields diverge on physical charge
D = Faradays laws tell us that
E = Btand Ampres law relate magnetic fields to currents by
H = JUniversit of California Berkele EECS 117 Lecture 17 . 29/
Are These Equations Complete?
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
30/33
Are These Equations Complete?
Are these equations complete and self-consistent? Inother words, do they over-specify the problem or aresome equations still missing? Furthermore, are they
self-consistent?Mathematics tells us that ( H) = 0, which impliesthat
J = 0But this can only hold for steady fields. In general, byconservation of charge we know that
J = t
Universit of California Berkele EECS 117 Lecture 17 . 30/
Maxwells Displacement Current
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
31/33
Maxwell s Displacement Current
In other words, we have to add something to the RHS ofAmpres eq. to make it self-consistent!
Maxwell was the first to make this observation. Since
D = , its natural to add a displacement current tothe Ampres eq.
H = J +D
t
This now makes our eq. self-consistent since
H = 0 = J + Dt
H = 0 = J + Dt = J + tUniversit of California Berkele EECS 117 Lecture 17 . 31/
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
32/33
Displacement Current of a Capacitor
-
8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law
33/33
Displacement Current of a Capacitor
The answer to this contradiction is displacementcurrent. If current is flowing in this circuit, then theelectric field between the capacitor plates must be
changing. ThusDt = 0
So the displacement current cutting surface S2 must bethe same as the conductive current cutting through
surface S1 S1
Jc dS =S2
D
t dS
Universit of California Berkele EECS 117 Lecture 17 . 33/