Find the Physics
SP212
Ch. 32 - Maxwell’s Laws
Maj Jeremy Best USMC
Physics Department, U.S. Naval Academy
March 30, 2016
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
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Maxwell’s Equations
March 30, 2016
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Gauss’s Law for Electric Fields
James Clerk Maxwell was a Scottish
physicist of the mid-XIXth Century
(19th). His contributions to human
knowledge rival those of Newton and
Einstein. In this course, we are most
concerned with his contributions to
electromagnetism. Which are
summarized in 4 vector differential
equations.
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
I
ΦE =
~E · d A
~ = qenc
0
The total flux of the electric field through a closed
surface is proportional to the net electric charge qenc
enclosed within that surface
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Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
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Solution:
Problem: Gauss’ Law for Electric Fields
A uniformly charged conducting sphere of diameter
1.2 m has a surface charge density of 8.1 µC /m2 . What
is the total electric flux leaving the surface of the sphere?
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
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d
qenc = σA A = 4πr 2
I2
~ = qenc
ΦE = ~E · d A
0
qenc
3.6644 × 10−5 C
ΦE =
=
C2
0
8.85 × 10−12 Nm
2
2
Nm
= 4.1386 × 106
=V ·m
C
r=
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
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Gauss’s Law for Magnetism
Problem: Flux
I
ΦB =
Along the flat top face, which has a
radius of 3.2 cm, a perpendicular field
~ of magnitude 0.17 T is directed
B
outward. Along the flat bottom face, a
magnetic flux of 0.65 mWb is directed
outward. What are the (a) magnitude
and (b) direction (inward or outward) of
the magnetic flux through the curved
part of the surface?
~ · dA
~ =0
B
This equation implies that there is no such thing as
magnetic monopoles, an isolated N or S pole. Magnets
only come in dipoles
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
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Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
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Faraday’s Law aka:Induced Electric Fields
Solution:
I
ΦB =
~ · dA
~ =0
B
I
ΦB top = BA = (.17T )(.0322 ∗ π)
ΦB bottom = BA = .65 × 10−3 Wb
Φin = ΦB top + ΦB bottom
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
~E · d~s = −dΦB
dt
A changing magnetic flux creates an induced electric
field.
March 30, 2016
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Problem: Faraday’s Law
A long straight wire is in the plane of a
rectangular conducting loop. The
straight wire carries a constant current
i, as shown. While the wire is being
moved toward the rectangle the
current in the rectangle is:
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
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Solution:
We have Faradays Law, Lenz’ Law, Biot-Savart and
Gauss’ Law for Magnetism all with this simple example...
I
~E · d~s = −dΦB
I dt
~ · dA
~ =0
ΦB = B
B=
µ0 i
2πR
So what happens? What does the field look like?
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
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Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
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3/4 of Maxwell’s equations were review, but the last one
included a new term. Compare:
I
~ · d~s = µ0 ienc
B
I
~ · d~s = µ0 ienc + µ0 0 dΦE
B
dt
Ampere-Maxwell Law
I
~ · d~s = µ0 ienc + µ0 0 dΦE
B
dt
A changing electric field induces an magnetic field, nicely
symmetric with the last equation we saw
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
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Displacement Current
Clearly, the quantity 0 (dΦE /dt) must have the
dimension of current. Appropriately, we call it the
displacement current, id .
I
~ · d~s = µ0 ienc + µ0 id,enc
B
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SP212
March 30, 2016
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Displacement Current
The classic example of the need for the displacement
current is seen while charging a capacitor.
How is the displacement current related to the “real”
current that is charging a capacitor? The electric field of
a capacitor is related to the charge on the capacitor:
q = 0 AE
dq
d(AE )
= ireal = 0
dt
dt
dΦE
ireal = 0
dt
ireal = id
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
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Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
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Solution:
Problem: Displacement Current
At what rate must the potential difference between the
plates of a parallel-plate capacitor with a 1.5 µF
capacitance be changed to produce a displacement
current of 1.9 A?
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
id = 0
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Maxwell’s Equations
March 30, 2016
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
March 30, 2016
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Problem: Maxwell Eqns: Which is False?
A. A magnetic field that changes in time creates an
electric field.
B. An electric field that changes in time creates a
magnetic field.
C. If the electric flux through a Gaussian surface is
negative, there is a net negative charge in the interior
volume bounded by the surface.
D. As a set, Maxwell’s equations imply the existence of
electromagnetic waves.
E. A positive magnetic flux can be obtained with a
Gaussian surface that surrounds just the north pole of a
magnet.
The most important Slide in this course. Seriously.
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
d(E )
dΦE
d(EA)
= A0
= 0
dt
dt
dt
V
0 A dV
E=
id =
d
d dt
dV
id d
id
=
=
dt
0 A C
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SP212
March 30, 2016
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