lecture 17 magnetic forcestorque, faraday’s law

8/2/2019 Lecture 17 Magnetic ForcesTorque, Faradays Law

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EECS 117

Lecture 17: Magnetic Forces/Torque, Faradays Law

Prof. Niknejad

University of California, Berkeley

Universit of California Berkele EECS 117 Lecture 17 . 1/


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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



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



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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



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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



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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/


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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



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/


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



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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



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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



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



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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



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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/


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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.


F d Bi Di


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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

=


F d L i Diff ti l F


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.


E l T f


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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


G ti S k !


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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


V t P t ti l


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=


M V t P t ti l


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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.


Is the vector potential real?


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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


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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)



24/33

Vector Potential Outside of Solenoid


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.


Circuit Application of Transformers


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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


Voltage Transformer


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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


Power Transmission


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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!


Summary So Far


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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?


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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


Maxwells Displacement Current


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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/


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Displacement Current of a Capacitor


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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


lecture 17 magnetic forcestorque, faraday’s law

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