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Physics 24 and theShakespearian Tragedy

Coulomb’s Law

Electric Fields

Electric Energy

Circuits

MagneticFields

Faraday’s Law, Maxwell’s Equations

Electromagnetic Waves

GeometricalOptics

WaveOptics

Final Exam

Today’s agenda:

Induced emf.You must understand how changing magnetic flux can induce an emf, and be able to determine the direction of the induced emf.

Faraday’s Law. You must be able to use Faraday’s Law to calculate the emf induced in a circuit.

Lenz’s Law.You must be able to use Lenz’s Law to determine the direction induced current, and therefore induced emf.

Generators.You must understand how generators work, and use Faraday’s Law to calculate numerical values of parameters associated with generators.

Back emf.You must be able to use Lenz’s law to explain back emf.

Induced emf and Faraday’s Law

Magnetic Induction

We have found that an electric current can give rise to a magnetic field…

I wonder if a magnetic field can somehow give rise to an electric current…

An electric current is induced if there is a closed circuit (e.g., loop of wire) in the changing magnetic flux.

It is observed experimentally that changes in magnetic flux induce an emf in a conductor.

A constant magnetic flux does not induce an emf—it takes a changing magnetic flux.

B

B

I

Note that “change” may or may not not require observable (to you) motion.

A magnet may move through a loop of wire

region of magnetic

fieldchange area of loop inside magnetic field

N S

rotate coil in magnetic field

this part of the loop is closest to your eyes

A magnet may move through a loop of wire, or a loop of wire may be moved through a magnetic field. These involve observable motion.

B IN S

move magnet toward coilv

I

A changing current in a loop of wire

In the this case, nothing observable (to your eye) is moving, although, of course microscopically, electrons are in motion.

Induced emf is produced by a changing magnetic flux.

changing I

changing B

induced I

A changing current in a loop of wire gives rise to a changing magnetic field (predicted by Ampere’s law)

A changing current in a loop of wire gives rise to a changing magnetic field (predicted by Ampere’s law) which can induce a current in another nearby loop of wire.

Today’s agenda:






We can quantify the induced emf described qualitatively in the previous section of this lecture by using magnetic flux.Experimentally, if the flux through N loops of wire changes by dB in a time dt, the induced emf is

Bd = - N .

dtε

Faraday’s law of induction is one of the fundamental laws of electricity and magnetism.

I wonder why the – sign…

*Faraday’s Law of Magnetic Induction

Your text, page 959 of the 14th edition, shows how to determine the direction of the induced emf. Argh! Lenz’s Law, coming soon, is much easier.

Baverage = - N .

tε

*Well, one expression of Faraday’s Law

This is another expression of Faraday’s Law:

Bd = - N ,

dtε

Bd

E ds= - dt

is the magnetic flux.

Faraday’s Law of Magnetic Induction

B B dA

We’ll use this version in the next

lecture.

In the equation

The fine print, put here for me to ponder, and not for students to worry about. A magnetic force does no work on a moving (or stationary) charged particle. Therefore the magnetic force cannot change a charged particle’s potential energy or electric potential. But electric fields can do work. This equation shows that a changing magnetic flux induces an electric field, which can change a charged particle’s potential energy. This induced electric field is responsible for induced emf. During this lecture, we are mostly going to examine how a changing magnetic flux induces emf, without concerning ourselves with the “middleman” induced electric field.

N SI

v

+ -

Example: move a magnet towards a coil of wire.

N=5 turns A=0.002 m2

dB = 0.4 T/s

dt

B dA

Bdd

= - N = - N dt dt

ε

d BA = - N (what assumptions did I make here?)

dtε

dB = - N A

dtε

2 T = - 5 0.002 m 0.4 =-0.004 V

sε

Possible homework hint: if B varies but loop B.

B Bd B dA B(t) dA

Ways to induce an emf:

change B

change the area of the loop in the field

Possible homework hint: for a circular loop, C=2R, so A=r2=(C/2)2=C2/4, so you can express d(BA)/dt in terms of dC/dt.

Possible Homework Hint.

I

The magnetic field is not uniform through the loop, so you can’t use BA to calculate the flux. Take an infinitesimally thin strip. Then the flux is d = BdAstrip. Integrate from a to b to get the flux through the strip.

a

b

changing current changes B through conducting loop

B

Ways to induce an emf (continued):

Ways to induce an emf (continued):

change the orientation of the loop in the field

=90 =45 =0

Today’s agenda:






Lenz’s law—An induced emf always gives rise to a current whose magnetic field opposes the change in flux.*

Experimentally…

*Think of the current resulting from the induced emf as “trying” to maintain the status quo—to prevent change.

N SI

v

+ -

If Lenz’s law were not true—if there were a + sign in Faraday’s law—then a changing magnetic field would produce a current, which would further increase the magnetic field, further increasing the current, making the magnetic field still bigger…

…violating conservation of energy and ripping apart the very fabric of the universe…

In which direction is the current induced in the coil for each situation shown? Practice on your own. In lecture, skip to here.

(counterclockwise) (no current)

Practice with Lenz’s Law.

(counterclockwise) (clockwise)

Rotating the coil about the vertical diameter by pulling the left side toward the reader and pushing the right side away from the reader in a magnetic field that points from right to left in the plane of the page.

(counterclockwise)

You can use Faraday’s Law (as written above) to calculate the magnitude of the emf (or whatever the problem wants). Then use Lenz’s Law to figure out the direction of the induced current (or the direction of whatever the problem wants).

Faraday’s LawBd

= - N dt

ε

The direction of the induced emf is in the direction of the current that flows in response to the flux change. We usually ask you to calculate the magnitude of the induced emf ( || ) and separately specify its direction.

Magnetic flux is not a vector. Like electrical current, it is a scalar. Just as we talk about current direction (even though it is not a vector), we often talk about flux direction (even though it is not a vector). Keep this in mind if your recitation instructor talks about the direction of magnetic flux.

Today’s agenda:






Skip to slide 27 (the skipped slides don’t add anything

new).

Induced emf: an overview

An emf is induced in a conductor moving in a magnetic field.Your text introduces four ways of producing induced emf.

Bd= -

dt

side view

BA

1. Flux change through a conducting loop produces an emf: rotating loop.

= NBA sin tε

NBAI = sin t

R

P= INBA sin t

start with this

derive these

2. Flux change through a conducting loop produces an emf:

Bd= -

dtv

ℓ

B

x=vdtdA

B

MF = I B

= B vεε B v

I = = R R

PP = F v =I Bv

start with these

derive these

2. Flux change through a conducting loop produces an emf: expanding loop.

3. Conductor moving in a magnetic field experiences an emf:

F= q E+v B

v

ℓ

B

B

–

+

= E (Mr. Ed)

start with these

derive this

ε = B v

This is called “motional” emf, because the emf results from the motion of the conductor in the magnetic field. The other three examples in this lecture are “induced” emf because they involve the change of magnetic flux through a current loop, as described by Faraday’s Law.

3. Conductor moving in a magnetic field experiences an emf: magnetic force on charged particles.

4. Flux change through a conducting loop produces an emf: moving loop.

Bd= -

dt

start with this

derive these

ε = B v

B v

I = P =I BvR

Let’s look in detail at each of these four ways of using flux change or motion to produce an emf.

Method 1…

Generators and Motors: a basic introduction

Take a loop of wire in a magnetic field and rotate it with an angular speed .

S N

side view

BA

B =B A = BA cos

Choose 0=0. Then

0= t = t .

B = BA cos t

Bd = -

dtε

Generators are an application of induced emf.

If there are N loops in the coil

Bd = - N

dtε

d BA cos t = - N

dtε

= NBA sin tε

side view

BA

|| is maximum when = t = 90° or 270°; i.e., when B is zero. The rate at which the magnetic flux is changing is then maximum. On the other hand, is zero when the magnetic flux is maximum.

The NBA equation!

emf, current and power from a generator

= NBA sin tε

NBAI = = sin t

R Rε

P= I = INBA sin tε

None of these are on your starting equation sheet!

= N B A ω sin ωt ε

max = N B A ωε

maxN = B A ωε

-2 2 -1

170 VN =

0.15 T 2×10 m 2 ×60 s

N = 150 (turns)

Example: the armature of a 60 Hz ac generator rotates in a 0.15 T magnetic field. If the area of the coil is 2x10-2 m2, how many loops must the coil contain if the peak output is to be max = 170 V?

“Legal” for me, because I just derived it (but not for

you)!

I will not work this in lecture. Please study it on your own! Know where the 2 comes from.


Method 2…

Recall that one of the ways to induce an emf is to change the area of the loop in the magnetic field. Let’s see how this works.

v

ℓ

B

vdtdA

A U-shaped conductor and a moveable conducting rod are placed in a magnetic field, as shown.The rod moves to the right with a constant speed v for a time dt.

B

Another Kind of Generator: A Slidewire Generator

x

The area through which flux passes increases by dA.

Bd = - N

dtε

d BA = 1

dtε

B dA =

dtε

dx

= B dt

ε

= B v . ε

B and v are vector magnitudes, so they are always +. Wire length is always +. You use Lenz’s law to get the direction of the current.

v

ℓ

B

vdtdA

B

The loop is perpendicular to the magnetic field, so the magnetic flux through the loop is B = = BA. Calculate the induced emf using Faraday’s law:

B dA

x

The induced emf causes current to flow in the loop.

v

ℓ

B

vdtdA

Magnetic flux inside the loop increases (more area).

System “wants” to make the flux stay the same, so the current gives rise to a field inside the loop into the plane of the paper (to counteract the “extra” flux).Clockwise current!

I

Direction of current?

x

What would happen if the bar were moved to the left?

As the bar moves through the magnetic field, it “feels” a force

ℓ

B

I

A constant pulling force, equal in magnitude and opposite in direction, must be applied to keep the bar moving with a constant velocity.

MF = I B

P MF = F = I B

v

FM FP

x

If the loop has resistance R, the current is

v

ℓ

B

(as expected).I

Power and current.

ε B vI = = .

R R

And the power is

PP = F v =I Bv

2P = I IR = I R

Mechanical energy (from the pulling force) has been converted into electrical energy, and the electrical energy is then dissipated by the resistance of the wire.

x

You might find it useful to look at Dr. Waddill’s lecture on Faraday’s Law, from several semesters back. Click here to view the lecture.

If the above link doesn’t work, try copying and pasting this into your browser address bar:http://campus.mst.edu/physics/courses/24/Handouts/Lec_18.ppt

http://campus.mst.edu/physics/courses/24/Handouts/Lec_18.ppt


Method 3…

Motional emf is the emf induced in a conductor moving in a magnetic field.

Example 3 of motional emf: moving conductor in B field.

v

ℓ

B

B

If a conductor (purple bar) moves with speed v in a magnetic field, the electrons in the bar experience a force

MF = qv B= - ev B

The force on the electrons is “up,” so the “top” end of the bar acquires a net – charge and the “bottom” end of the bar acquires a net + charge. The charges in the bar are “separated.”

“up”

–

+

This is a simplified version of reality, but it gives you the right “feel.”

The separated charges in the bar produce an electric field pointing “up” the bar. The emf across the length of the bar is

v

ℓ

B

B

The electric field exerts a “downward” force on the electrons:

EF = qE= - eE

An equilibrium condition is reached, where the magnetic and electric forces are equal in magnitude and opposite in direction.

“up”

–

+

ε

evB= eE= e

ε = E

ε = B v

E

The separated charges in the bar produce an electric field pointing “up” the bar. The emf across the length of the bar is

Homework Hint!

v

ℓeffective

B

B

When the moving conductor is “tilted” relative to its direction of motion (when is not perpen-dicular to the conductor), you must use the “effective length” of the conductor.*

–

+

Caution: you do not have to use this technique for homework problems 29.25 or 29.28 (if assigned this semester). Be sure to understand why!

V E E cos

effectiveE ℓeffective = ℓ sin if you use the if you define as the angle relative to the

horizontal

v

*This is because the magnetic force is , and not directed along the conductor. Let’s not worry about showing this.


Method 4…

Example 4 of induced emf: flux change through conducting loop. (Entire loop is moving.)

I’ll include some numbers with this example.

Remember, it’s the flux change that induces the emf. Flux has no direction associated with it. However, the presence of flux is due to the presence of a magnetic field, which does have a direction, and allows us to use Lenz’s law to determine the “direction” of current and emf.

B = 0.6 T

5 c

mA square coil of side 5 cm contains 100 loops and is positioned perpendicular to a uniform 0.6 T magnetic field. It is quickly and uniformly pulled from the field (moving to B) to a region where the field drops abruptly to zero. It takes 0.10 s to remove the coil, whose resistance is 100 .

Final: Bf = 0 .

B = Bf - Bi = 0 - BA = - (0.6 T) (0.05 m)2 = - 1.5x10-3 Wb.

(a) Find the change in flux through the coil.

Initial: Bi = = BA .B dA

The current must flow clockwise to induce an “inward” magnetic field (which replaces the “removed” magnetic field).

initialfinal

(b) Find the current and emf induced.

Current will begin to flow when the coil starts to exit the magnetic field. Because of the resistance of the coil, the current will eventually stop flowing after the coil has left the magnetic field.

The induced emf is

Bd d(BA) dA = - N = - N = - N B

dt dt dtε

v

dA

d xdA dx = = = v

dt dt dt

x

x

m = 100 0.6 T 0.05 m 0.5

sε

= 1.5 Vε

“uniformly” pulled

= - N B vε

x 5 cm m

v = = = 0.5 t 0.1 s s

The induced current is

I = = = 15 mA .R 100 Ωε 1.5 V

Current flows “only*” during the time flux changes.

E = P t = I2R t = (1.5x10-2 A)2 (100 ) (0.1 s) = 2.25x10-3 J

*Remember: if there were no resistance in the loop, the current would flow indefinitely. However, the resistance quickly halts the flow of current once the magnetic flux stops changing.

The loop has to be “pulled” out of the magnetic field, so there is a pulling force, which does work.

The “pulling” force is opposed by a magnetic force on the current flowing in the wire. If the loop is pulled “uniformly” out of the magnetic field (no acceleration) the pulling and magnetic forces are equal in magnitude and opposite in direction.

(c) How much energy is dissipated in the coil?

(d) Discuss the forces involved in this example.

The flux change occurs only when the coil is in the process of leaving the region of magnetic field.

No flux change.No emf.No current.No work (why?).

Flux changes. emf induced. Current flows. Work (equal to FD) is done.

D

Fapplied

No flux change. No emf. No current. (No work.)

Remember, a force is needed only when the coil is partly in the field region.

(e) Calculate the force necessary to pull the coil from the field.

I

magF = N IL B

where L is a vector in the direction of I having a magnitude equal to the length of the wire inside the field region.

I

L1

L2 L3

F1

F2

F3

Sorry about the “busy” slide!The forces should be shown acting at the centers of the coil sides in the field.

There must be a pulling force to the right to overcome the net magnetic force to the left.

Fpull

L4=0

Multiply by N because there are N loops in the coil.

-2 -2 -2mag pullF = NILB= 100 1.5 10 5 10 0.6 =4.5 10 N=F

Magnitudes only (direction shown in diagram):

I

L1

L2 L3

F1

F2

F3

Fpull

This calculation assumes the coil is pulled out “uniformly;” i.e., no acceleration, so Fpull = Fmag.

ˆ ˆ

pull pullW =F D = NILB i L i

Work done by pulling force:

IFpull

D

-2 -2 -2 -3pullW = 100 1.5 10 5 10 0.6 5 10 =2.25 10 J

x

The work done by the pulling force is equal to the electrical energy provided to (and dissipated in) the coil.

Concluding Remarks

I will post a PowerPoint illustrating the operation of motors, discussing back emf, and giving two more examples of motional emf calculations. You can find the PowerPoint in the “resources” section for this lecture on the course web site.These slides (posted under “resources”) are intended to supplement your reading. They are optional. I will not post a narration of their content.

Now that you are an expert at designing generators, I thought it wise to direct you towards some handy gadgets you can use if you need to debug your non-working generator.

If you build a generator and it doesn’t seem to be working…

Handy Hint: Debugging Your Homemade Generator

…or if you want to test a wall socket to see if it is “live…”

…simply purchase a Vilcus Plug Dactyloadapter from ThinkGeek ( http://www.thinkgeek.com/stuff/41/lebedev.shtml )

and follow the instructions.

European model shown. US model also available.

http://www.thinkgeek.com/stuff/41/lebedev.shtml

http://www.thinkgeek.com/stuff/41/lebedev.shtml

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Documents

induced emf

lenzs law

changing magnetic flux

changing magnetic field

amperes law

changing current

faradays law of induction

magnetic fieldthis