Download Lecture 16 - The Local Group

Lecture 16
Maxwell’s Equations
Announcements
— Email me by 11:59pm today, if you want a
left-handed seat for the final exam.
— Seating chart to be distributed via EEE.
http://www.classrooms.uci.edu/gac/SSPA1100.html
Ampere’s Law and the “displacement” current
Recall Ampere’s law as we learned it last chapter:
!" "
#∫ B • dl = µ0 I encl
It says that the line integral of B around the boundary of a surface is
proportional to the current passing through that surface.
- We are free to choose any surface for a given boundary
This version of Ampere’s Law is only valid for static charge distributions
and steady currents!
Ampere’s law does NOT work for this surface.
A problem with Ampere’s Law!
Consider a capacitor being
charged by a wire with current iC .
1. Draw a circular path around the wire
& use the area of a flat circle in
Ampere’s law. For this surface:
Ienc = iC
2. Now use a “bulging” surface which
extends into the region between the
capacitor plates, but is bounded by the
same path as the plane surface.
◦ For this (bulging) surface, Ienc = 0
• Surface 1:
• Surface 2:
! Clear contradiction !
Displacement Current
To resolve this, Maxwell noted that the electric field
is changing through the bulging surface.
•For a parallel plate capacitor:
q = Cv = CEd = ε A Ed = ε AE = εΦ E
d
•As the capacitor charges, the electric field between
the capacitor plates increases and so does the electric
flux through the bulging surface:
dq
dΦ E
iC =
dt
=ε
dt
Maxwell invented a fictitious displacement current
between the plates:
dΦ E
iD = ε
dt
As such, Ampere’s law becomes:
Corrected Ampere’s Law
dΦ E
iD = ε
dt
Maxwell resolved the problem by guessing that a changing
E-field produces a B field, just as the electric current does.
The changing E-field is equivalent to an electric current: iD
Displacement Current
The “displacement current” is attributed to the
change in the electric field between plates, and
treated like a “real” (conduction) current for
purposes of generating a magnetic field.
•In particular, a (displacement) current between the
capacitor plates should produce a circulating B field
between the plates, the same as for a current in a
straight wire.
Using Ampere’s law for a circle of radius r (r < R)
dΦ E
iD = ε
dt
Notice direction of induced B.
It is the same as for B around a wire !
Total current enclosed is jD ✕ area:
µ0 r
B=
i
2 C
2π R
and this, we can check
experimentally!
⎛ iD ⎞ 2
⎜⎝ 2 ⎟⎠ πr
πR
For a charging capacitor: iC=iD
Implications
When the experiment we’ve been thinking about with the charging capacitor is
actually performed, the predicted magnetic field between the plates is, in fact,
observed. - Maxwell’s theoretical hunch was right!
To summarize what it means:
•a changing magnetic field creates an induced electric field (Faraday’s Law)
•a changing electric field also creates a magnetic field
With the generalized version of Ampere’s Law, including displacement currents
due to a changing electric field, Maxwell was able to unify the laws of electricity
and magnetism into a single set of 4 equations.
•Maxwell’s equations predicted, among other things, that disturbances in the E and
B fields – an “electromagnetic wave” – could propagate through space and the
speed of these waves is exactly the speed of light.
Maxwell’s equations
•Gauss’s law for electricity:
Relate charge distribution and the electric field
•Gauss’s law for magnetism:
Magnetic field
(no magnetic monopoles)
•Faraday’s law of induction:
Electrical effect of a changing
magnetic field
•Ampere’s law:
Magnetic effect of moving charge
or a change in electrical field
dΦ
i =ε
According to Maxwell’s equations:
dt
• A point charge at rest produces E but not B.
• A point charge moving with a constant speed produces E & B.
• For a point charge to produce and EM wave, the charge must accelerate.
D
0
E
Just a bit more induction from Chapter 30….
Induced emf
Coil 1
Coil 2
i1
Two coils. Coil 1 and Coil 2.
- Coil 1 has a current i1 that is changing with time.
Induced emf
Coil 1
B
Coil 2
i1
Two coils. Coil 1 and Coil 2.
- Coil 1 has a current i1 that is changing with time.
- This produces a magnetic field B
B / i1
Induced emf
Coil 1
B
Coil 2
i1
Two coils. Coil 1 and Coil 2.
- Coil 1 has a current i1 that is changing with time.
- This produces a magnetic field B
B / i1
- The current is changing so the B-field changes:
dB
di1
/
dt
dt
Induced emf
Coil 1
Two coils. Coil 1 and Coil 2.
- Coil 1 has a current i1 that is changing with time.
- This produces a magnetic field B
B
Coil 2
i1
B / i1
ɸ
A2
- The current is changing so the B-field changes:
dB
di1
/
dt
dt
This will result in a changing magnetic flux through coil 2 and therefore an emf in coil 2:
dΦ B
ε=−
dt
E2 =
d
B2
dt
'
dB
di1
A2
/
dt
dt
Mutual inductance
Coil 1
We just saw that a changing current through coil 1
will result in an emf in coil 2:
B
Coil 2
i1
ε
E2 =
d
B2
dt
di1
/
dt
The constant of proportionality will depend on the
geometry of the two coils. Let’s call it M21:
E2 =
We call M21 the “Mutual
inductance” of the two
coils
di1
M21
dt
Mutual inductance
Coil 1
B
Coil 2
We just saw that a changing current through coil 1
will result in an emf in coil 2:
E2 =
i1
d
B2
dt
di1
/
dt
The constant of proportionality will depend on the
geometry of the two coils. Let’s call it M21:
E2 =
The definition of mutual
inductance can also be
written as
di1
M21
dt
M21 =
B2
i1
Mutual inductance: multiple loops
Coil 1
B
N2 turns
Coil 2
If there are multiple coils, each one will get an
induced emf, resulting in a total emf:
E2 =
i1
N2
d
B2
dt
di1
/
dt
The constant of proportionality will depend on the
geometry of the two sets of coils. Let’s call it M21:
E2 =
The definition of mutual
inductance can also be
written as
di1
M21
dt
M21
N2 B2
=
i1
Mutual Inductance
N1 turns
Coil 1
B
At the same time, any current in Coil 2 will also
induce an emf in Coil 1.
N2 turns
Coil 2
i2
Mutual Inductance
N1 turns
Coil 1
B
Remarkably it turns out that M12 is always equal
to M21:
N2 turns
Coil 2
M12 = M21 = M
i2
E1 =
E2 =
di2
M
dt
di1
M
dt
Mutual inductance
di2
di1
ε1 = −M
ε 2 = −M
dt
dt
Φ B1
Φ B2
M = N1
= N2
i2
i1
Remember: emf is induced in a circuit
ONLY if current in the other CHANGES.
➡The mutual inductance M tells us how
much emf results in one circuit from a
given change of current in the other.
➡M is a constant and depends on size,
shape, number of turns and relative
positions of the two coils.
- e.g. the coils are closer, more line of
flux will go through & M is greater.
M = N1
Φ B1
Φ
= N2 B2
i2
i1
Units of inductance
In SI units, Inductance is measured in a unit called a “henry” (H).
Recall that the SI unit of magnetic flux is the Weber, and inductance measures the
magnetic flux per unit of current. Thus
1 H = 1 Wb/A = 1 Vs/A = 1 Ωs = 1 J/A 2
One henry is a very large amount of inductance.
Typical values are usually measured in millihenry (mH) or microhenry (µH).
Pros and cons of mutual inductance
Can you think of any pros or cons of mutual inductance?
Pros and cons of mutual inductance
Mutual inductance can be a problem in the design of sensitive circuits,
since the changes of current in one part of a circuit can create emfs (and
currents) in a different part of the circuit.
On the other hand, mutual inductance can be used to couple two circuits
without an actual electrical connection, as in the electric toothbrush
charger.
•A coil in the charger is coaxial with a coil in the toothbrush.
•Changes in the (A/C) current in the charger’s coil induce an emf in the toothbrush’s coil that recharges the toothbrush’s battery.
•No electrical connection means no risk of electrocution when used
near water.
l=0.5m
A=10cm2=10-3m2
N1=1000
N2=10
Example: mutual
inductance of
tesla coil
The “Tesla coil” is one solenoid (N2 turns) wrapped around another
solenoid with N1 turns, length l and cross-sectional area A. No direct
connection between the two coils. Find the mutual inductance and
the emf in the inner coil when the current in the outer loop varies as
(
)
i2 (t) = 2.0 ×10 6 [A / s] t
di2
ε1 = −M
dt
Need to find M first.
l=0.5m
A=10cm2=10-3m2
N1=1000
N2=10
Example: mutual
inductance of
tesla coil
The “Tesla coil” is one solenoid (N2 turns) wrapped around another
solenoid with N1 turns, length l and cross-sectional area A. No direct
connection between the two coils. Find the mutual inductance and
the emf in the inner coil when the current in the outer loop varies as
M = N1
Φ B1
Φ
= N2 B2
i2
i1
(
)
i2 (t) = 2.0 ×10 6 [A / s] t
l=0.5m
A=10cm2=10-3m2
N1=1000
N2=10
Example: mutual
inductance of
tesla coil
The “Tesla coil” is one solenoid (N2 turns) wrapped around another
solenoid with N1 turns, length l and cross-sectional area A. No direct
connection between the two coils. Find the mutual inductance and
the emf in the inner coil when the current in the outer loop varies as
M = N1
Φ B1
Φ
= N2 B2
i2
i1
Mi1 = N2 Φ B2
Φ B2 = B1 A
B1 =
µ0 N1i1
l
(
)
i2 (t) = 2.0 ×10 6 [A / s] t
l=0.5m
A=10cm2=10-3m2
N1=1000
N2=10
Example: mutual
inductance of
tesla coil
The “Tesla coil” is one solenoid (N2 turns) wrapped around another
solenoid with N1 turns, length l and cross-sectional area A. No direct
connection between the two coils. Find the mutual inductance and
the emf in the inner coil when the current in the outer loop varies as
M = N1
Φ B1
Φ
= N2 B2
i2
i1
Mi1 = N2 Φ B2
Φ B2 = B1 A
µ Ni
B1 = 0 1 1
l
(
)
i2 (t) = 2.0 ×10 6 [A / s] t
µ0 N1i1
A = Mi1
l
N1 N2
M = µ0 A
l
N2 Φ B2 = N2
M = 4π ×10 −7 [Wb/Am ]×10 −3 ⎡⎣ m2 ⎤⎦
1000 ×10
= 25µH
0.5[m]
l=0.5m
A=10cm2=10-3m2
N1=1000
N2=10
Example: mutual
inductance of
tesla coil
The “Tesla coil” is one solenoid (N2 turns) wrapped around another
solenoid with N1 turns, length l and cross-sectional area A. No direct
connection between the two coils. Find the mutual inductance and
the emf in the inner coil when the current in the outer loop varies as
(
)
i2 (t) = 2.0 ×10 6 [A / s] t
di2
= 2.0 ×10 6 [A / s]
dt
di2
ε1 = −M
dt
ε1 = −25 ×10
−6
6
[H ]× 2 ×10 [A/s] = −50V
M from previous page
Self Inductance
Now consider a single coil, carrying a current i.
The current in the coil creates a solenoidal magnetic field inside the (same) coil.
The total magnetic flux through the coil will be proportional to the current in the (same)
coil:
NΦ B = Li
where ΦB is the flux through a single turn.
The constant
NΦ B
L=
i
is called the self-inductance of the coil.
L depends on the size and geometry of the coil,
but not the current.
Self -Induced emf
If the current in the coil is changing, then the flux will also change by an amount
proportional to the self-inductance:
dΦ B
di
=L
dt
dt
Faraday’s Law says that a changing flux creates an induced emf:
N
ε = −N
dΦ B
dt
ε = −L
di
dt
Thanks to the minus sign in Faraday’s Law (or equivalently, thanks to Lenz’s Law), the
induced emf will oppose the change that caused it.
•The self-induced emf in the coil will oppose the change in current that caused it.
•
If I increases, the emf induced will oppose the increase
•To measure unknown self-inductance, change current at know rate di/dt & measure emf to get L.
!Self-induction: The act of a changing current through a loop inducing an opposite current in that
same loop.
Self-inductance has the same units as mutual inductance and is also measured in henry (H)
L is called the self-inductance of the loop.
Inductance is a purely geometric quantity
Final Exam
Exam will be at 8am on Tuesday, March 21st in this room.
Exam will be cumulative.
Use of calculators will be permitted.
Bring pen/pencil and ID.
Show up on time (i.e. before 8am).
— Email me by 11:59pm today, if you want a
left-handed seat for the final exam.
— Seating chart to be distributed via EEE.
http://www.classrooms.uci.edu/gac/SSPA1100.html
Yes, there will be an equation sheet…