Lecture 19 Faraday’s Law and Maxwell’s Eq.

8/2/2019 Lecture 19 Faradays Law and Maxwells Eq.

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EECS 117 Lecture 19: Faradays Law and Maxwells Eq.

Prof. Niknejad

University of California, Berkeley

Universit of California Berkele EECS 117 Lecture 19 . 1/


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Magnetic Energy of a Circuit

Last lecture we derived that the total magnetic energy ina circuit is given by

E m = 12L1 I 21 + 12L

2 I 22 + MI 1 I 2

We would like to show that this implies that M

L1 L2 .

Lets re-write the above into the following positivedenite form

E m =12L1 I 1 +

M L1 I 2

2

+12 L2

M 2

L1 I 22

An important observation is that regardless of the

current I 1 or I 2 , the magnetic energy is non-negative, soE m 0Universit of California Berkele EECS 117 Lecture 19 . 2/


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Magnetic Energy is Always Positive

Consider the current I 2 = L 1M I 1 , which cancels the rstterm in E m

E m = 12

L2 M 2

L1I 22 0

Since I 2

2 0, we haveL2

M 2

L1 0Therefore its now clear that this implies

L1 L2

M 2



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Coupling Coefcient

Usually we express this inequality as

M = k

L1 L2

Where k is the coupling coefcient between twocircuits, with |k| 1.If two circuits are perfectly coupled (all ux from circuitone crosses circuit 2), k = 1 (ideal transformer)Note that M < 0 implies that k < 0, which is totallyreasonable as long as

klies on the unit interval

1 k 1Negative coupling just means that the ux gets invertedbefore crossing the second circuit. This is easilyachieved by winding the circuits with oppositeorientation.



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Motion (cont)

++

++

++

F e

The induced voltage in the bar

V ind = 2

1E d =

2

1(v B ) d



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A Moving Metal Bar

A

B

C

D

v 0B

R L

I

I

x

y

z

Consider the following generator. A bar of length

moves to the right with velocity v0 (always makingcontact with the rest of the circuit)

V ind = C (v B ) d = D

C (x v0 z B0 ) y dy = v0 B0

Current I ows in a direction to decrease the ux

(Lenzs Law)



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Energy Dissipated by R

This current ows through the resistor RL where theenergy of motion of the bar is converted to heat

The load will dissipate energy

P L = I 2 RL =(v0 B0 )2

RL

This power comes from the mechanical work in movingthe bar. The force experienced by a current carryingwire dF = Id

B

F m = I D

C dB = I

D

C y dy z B0 = IB 0

Thus P in = F m v0 = IB 0 v0 = P LUniversit of California Berkele EECS 117 Lecture 19 . 8/


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(im)Practical Example

Lets say we do this experiment using the earthsmagnetic eld

Use a bar with length = 1 m , B0 = 0 .5 GTo induce only 1 V, we have to move the bar at a speedof

v0 =V indB0 = 2 10

4

m/ sThe magnetic eld on the surface of a neutron star isabout B0

1012 G, or about 108 T . Even moving at a

speed of v0 = 1 m/ s, we generate

V ind = 10 8 V

Energy generation on a neutron star is easy!



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Back to Faradays Equation

Note that this problem is just as easy to solve usingFaradays Law

The ux crossing the loop is increasing at a constantrate

(t) = 0 + v0 tB 0

Where 0 is the initial ux at t = 0The induced voltage is simply

V ind

= d

dt = v0

B0



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An AC Generator

B

sinusoidal motion

+

V

If we connect our metal bar to a piston, in turnconnecting to a water-wheel or otherwise rotatingwheel, we have a crude generator

To generate substantial voltage, we need a strongmagnetic eld

Say we rotate the wheel at a rate of = 2 103 s 1 , or

1000 RPS (revolutions per second)



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An AC Generator (cont)

The ux is now

= 0 + B0 Am cos tWhere Am is the amplitude of oscillation. Taking thetime derivative

= Am B0 sin t = V indPlugging in some numbers, we see that with a relativelystrong magnetic eld of 1 T, an amplitude Am = 1 m , = 1 m , the voltage generated is reasonable

V ind = 2 103 sin t

The voltage is sinusoidal with frequency equal to therotation frequency



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AC Motor/Generator

I

B 0 = x B 0

I

A simple AC motor/generator consists of a rotating loopcutting through a constant magnetic eld. The slip rings

maintain contact with the loop as it rotates.



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AC Motor/Generator

If AC current is passed through the loop, it rotates at arate determined by the frequency. If, on the other hand,the loop is rotated mechanically and the circuit is closed

with a load, mechanical power is converted to electricityThe ux in the loop of area A is simply

= AB0

cos The phase = 0 t so

V ind = = AB 0 0 sin 0 tThis result can also be derived by using F m = qv B



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Maxwells Eq (Integral Form)

We have now studied the complete set of MaxwellsEquations . In Integral form

C E d = ddt S B dS

C H d =ddt S D dS + S J dS S D dS = V dV

S B dS = 0



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Maxwells Eq (Differential Form)

The elds are related by the following materialparameters

B = H D = E J = EFor most materials we assume that these are scalarrelations.

E = Bt

B = 0 0 Et + 0 (J +

P t + M )

E =10

( P )

B = 0Universit of California Berkele EECS 117 Lecture 19 . 16/


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Source Free Regions

In source free regions = 0 and J = 0 . Assume thematerial is uniform (no bound charges or currents)

E = Bt

B = 0 0

E

t

E = 0

B = 0



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Wave Motion

B

E E E

BB

wave motion

We can see intuitively that E t Bt

E t . . ., that

wave motion is possible


i i ll


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Time Harmonic Maxwells Eq.

Under time-harmonic conditions (many importantpractical cases are time harmonic, or nearly so, or elseFourier analysis can handle non-harmonic cases)

E = jB

H = J + jD

E = B = 0

These equations are not all independent. Take thedivergence of the curl, for instance

(

E )

0 =

j

B


H i E ( )


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Harmonic Eq. (cont)

In other words, the non-existence of magnetic charge isbuilt-in to our curl equation. If magnetic charge is everobserved, wed have to modify our equations

This is analogous to the displacement current thatMaxwell introduced to make the curl of H equationself-consistent

(H ) 0 = J + j D

J =

t=

j

This implies that D = , so Gauss law is built-in toour curl equations as well.


T i l B d C di i


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Tangential Boundary Conditions

The boundary conditions on the E-eld at the interfaceof two media is

n (E 1 E 2 ) = 0Or equivalently, E 1 t = E 2 t . If magnetic charges are everfound, then this condition will have to include the

possibility of a surface magnetic currentThe boundary conditions on H are similar

n (H1

H2

) = Js

For the interface of a perfect conductor, for example, asurface current ows so that ( H 2 = 0 )

n H 1 = J sUniversit of California Berkele EECS 117 Lecture 19 . 21/

B d C diti f C t


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Boundary Conditions for Current

Applying the pillbox argument to the divergence ofcurrent

V ( J )dV = S J dS = V t dV

in the limit

J 1 n J 2 n = stwhere s is the surface current. In the static case

J 1 n = J 2 n

implies that 1 E 1 = 2 E 2 . This implies that s = 0 since

1

E 1

= 2

E 2

(unless the ratios of match the ratio of perfectly!)Universit of California Berkele EECS 117 Lecture 19 22/

Lecture 19 Faraday’s Law and Maxwell’s Eq.

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Transcript of Lecture 19 Faraday’s Law and Maxwell’s Eq.