UNIT 10: MAGNETISMAP Physics C

AP PHYSICS C

UNIT 10: MAGNETISM

PART 1

MAGNETIC FIELDS AND

FORCES

Facts about Magnetism

• Magnets have 2 poles (north

and south)

• Like poles repel

• Unlike poles attract

• Magnets create a

MAGNETIC FIELD around

them

Magnetic Field• A bar magnet has a magnetic field

around it. This field is 3D in nature and often represented by lines LEAVING north and ENTERING south

• To define a magnetic field you need to understand the MAGNITUDE and DIRECTION

• We sometimes call the magnetic field a B-Field as the letter “B” is the SYMBOL for a magnetic field.

• Magnetic field strength is measured with the TESLA (T) as the unit.

• B fields can also be measured in Gauss (G). 1 T = 10,000 G

Magnetic Force

• The bar magnets do not directly exert a force on one another, but it is the magnetic fields that exert the forces.

• When magnetic field lines are in the opposite direction they attract.

• When the lines are in the same direction they repel.

Repel

Attract

Earth is a magnet too!

The magnetic north pole of the

EARTH corresponds with geographic

south and vice versa.

So when you use a compass the NORTH

POLE of the compass must be attracted to

a South Pole on the earth if you wanted to

travel north.

This magnetic field is very important in

that it prevents the earth from being

bombarded from high energy particles.

This key to this protection is that the

particles MUST be moving!

Magnetic Force on a moving charge• If a MOVING CHARGE moves into

a magnetic field it will experience a

MAGNETIC FORCE. This

deflection is 3D in nature.

• The conditions for the force are:

• Must have a magnetic field present

• Charge must be moving

• Charge must be positive or negative

• Charge must be moving

PERPENDICULAR to the field.

sinqvBF

BvqF

B

B

N

N

S

S

-

vo

B

Example

B

B

B

F

xxxF

qvBF

)1055)(100.1)(106.1(

190sin,90,

6519

A proton moves with a speed of 1.0x105 m/s through the Earth’s magnetic field, which has

a value of 55mT at a particular location. When the proton moves eastward, the magnetic

force is a maximum, and when it moves northward, no magnetic force acts upon it. What

is the magnitude and direction of the maximum magnetic force acting on the proton?

8.8x10-19 N

The direction cannot be determined precisely by the given information. Since

no force acts on the proton when it moves northward (meaning the angle is

equal to ZERO), we can infer that the magnetic field must either go northward

or southward.

Direction of the magnetic force?

Right Hand RuleTo determine the DIRECTION of the

force on a POSITIVE charge we use a special technique that helps us understand the 3D/perpendicular nature of magnetic fields.

•The Pointer = Direction of velocity or current

•The Middle Finger = Direction of the B field

•The Thumb = Direction of the Force

For NEGATIVE charges use left hand!

ExampleDetermine the direction of the unknown variable for a proton

moving in the field using the coordinate axis given

+y

+x+z

B = -x

v = +y

F =+z

B =+Z

v = +x

F =-y

B = -z

v = +y

F =-x

ExampleDetermine the direction of the unknown variable for an

electron using the coordinate axis given.

+y

+x+z

B = +x

v = +y

F =+z

B =

v = - x

F = +y

-z

F

B

B = +z

v =

F = +y+x

Motion due to a Magnetic Force

• When a charged particle moves in a

magnetic field it experiences a force that

is perpendicular to the velocity

• Since the force is perpendicular to the

velocity, the charged particle experiences

an acceleration that is perpendicular to

the velocity

• The magnitude of the velocity does not

change, but the direction of the velocity

does producing circular motion

• The magnetic force does no work on the

particle (it does not gain or lose energy).

Magnetic Force and Circular Motion

X X X X X X X X X X X X

X X X X X X X X X X X X

X X X X X X X X X X X X

X X X X X X X X X X X X

X X X X X X X X X X X X

vB

- -

FB

FB

FB

FB -

-

-

Suppose we have an electron

traveling at a velocity , v, entering

a magnetic field, B, directed into

the page. What happens after the

initial force acts on the charge?

Remember that the magnetic field

only changes the direction of the

charge, not the velocity. (the B

field does no work)

Magnetic Force and Circular Motion

•The magnetic force is equal to the

centripetal force and thus can be used

to solve for the circular path.

•If the radius is known, could be used

to solve for the MASS of the ion. This

could be used to determine the

material of the object. This technique

is used in a Mass Spectrometer.

There are many “other” types of

forces that can be set equal to the

magnetic force.

Example

?

5.0

250

105.2

106.1

26

19

r

TB

VV

kgxm

Cxq

qB

mvr

r

mvqvBFF cB

2

26

19

2

105.2

)106.1)(250(22

21

x

x

m

Vqv

q

mv

q

K

q

WV

)5.0)(106.1(

)568,56)(105.2(19

26

x

xr

A singly charged positive ion has a mass of 2.5 x 10-26 kg. After being

accelerated through a potential difference of 250 V, the ion enters a

magnetic field of 0.5 T, in a direction perpendicular to the field.

Calculate the radius of the path of the ion in the field.

We need to

solve for the

velocity!

56,568 m/s

0.0177 m

A Charge in Crossed Electric and Magnetic Fields

• Suppose a proton is injected

between two parallel plates

containing an electric field E in the –y

direction and a magnetic field B in

the –z direction.

• What is the direction of the forces on

the proton from each field?

• The force equation can be rewritten

as

• This is known as the Lorentz Force Law

• Under what conditions will the charge

not undergo a change in direction?• When F = 0, so E = -vxB

)( BvEqF

Velocity Selector

• If the particle is positively

charged then the magnetic

force on the particle will be

downwards and the electric

force will be upwards.

• If the velocity of the charged particle is just right then the net

force on the charged particle will be zero.

E

B.v

q

BF

EF

B

EvqEqvB

Charges Moving in a WireThe charges could be moving together in a wire. Thus, if the wire had a

CURRENT (moving charges), it too will experience a force when placed in a magnetic field.

The same right hand rule can

be used to find the direction

of the force on the wire.

Charges Moving in a Wire

BIlF

IlBF

lengthldxIdxBF

dt

dqIdxB

dt

dqB

dt

dxdqF

dt

dxvdqvBqvBF

B

B

B

B

B

sin

sin

sin

sinsin

The formula for the force acting on a

current-carrying wire in an EXTERNAL

B field can be derived from the force

acting on a single charge in the wire.

Remember this is CONVENTIONAL

current!

Example

B

B

B

F

xF

ILBF

90sin)1050.0)(36)(22(

sin

4

A 36-m length wire carries a current

of 22A running from right to left.

Calculate the magnitude and

direction of the magnetic force

acting on the wire if it is placed in a

magnetic field with a magnitude of

0.50 x10-4 T and directed up the

page.

0.0396 N

B = +y

I = -x

F =

+y

+x+z

-z, into the page

A Current-Carrying Loop in a B Field

• A loop of a current-carrying conductor is

placed in an external magnetic field that is

directed to the left.

• Describe the forces on each side of the

loop.

• The left side will be pushed outward.

• The right side will be pushed inward.

• The top and bottom will feel no forces.

• This means that the loop will begin to

ROTATE because the force has supplied

a TORQUE.

• This is the basic principle of how electric

motors and galvanometers work.

The Electric Motor

Galvanometer

• A device that is used to measure

electric current.

• Electric current passes through the

coil which then experiences a force

caused by the magnetic field it is

placed in. A spring is attached to

counter the force.

• The scale is calibrated to correspond

to different currents or voltages.

Torque on a Rectangular Current-Carrying Loop

B

B

BIA

BWLI

ILBW

Fr

B

B

B

B

B

m

m

sin

sin)(

sin)(

sin))(2/(2

Where μ is called the “magnetic

moment”

IAIWLIL 2m

We will most often use the

equation in the form

where N is the number of loops

sinNBIAB

Example

A rectangular shaped loop with a length of 10 cm and a width of 7 cm has a pivot placed

along its length through the center. The loop experiences an external magnetic field of

0.25 Tesla.

a) How much torque is exerted on the loop when it has a current of 0.85 A and is

perpendicular to the field?

b) How much torque is exerted on the loop when it has a current of 2.7 A and is at an

angle of 18o to the field?

The Hall Effect

• Another effect occurs when a wire

(or other conductor) carrying a

current is immersed in a

magnetic field.

• Consider a rectangular conductor with

conventional current I running across it immersed in a

perpendicular magnetic field.

• The positive charges will experience a magnetic force that

will cause them to move toward the edge of the conductor,

leaving an apparent negative charge on the opposite edge.

• This produces a potential difference across the conductor.

The Hall Effect

• The Hall Effect can be used to

determine the sign of charges that

actually make up the current.

• If the charges are positive, then

potential will be higher at the upper

edge.

• If charges are negative, then potential

will be higher at the lower edge.

• Experiment shows that the second

case is true.

• The charge carriers are in fact negative

electrons.

AP PHYSICS C

UNIT 10: MAGNETISM

PART 2

MAGNETIC SOURCES

Causes of magnetic forceEarlier we stated that a current-carrying wire in a magnetic field

experiences a force that causes it to move, but WHY does the wire move? It is easy to say the EXTERNAL field moved it. But how can an external magnetic field FORCE the wire to move in a certain direction?

THE WIRE ITSELF MUST BE MAGNETIC!!! The wire has its own

INTERNAL MAGNETIC FIELD that is attracted or repulsed by the

EXTERNAL FIELD.

As it turns out, the wire’s OWN internal magnetic

field makes concentric circles round the wire.

A current carrying wire’s INTERNAL magnetic

field

To figure out the DIRECTION of this INTERNAL field you use the right hand rule. You point your thumb in the direction of the current then CURL your fingers. Your fingers will point in the direction of the magnetic field

The MAGNITUDE of the internal field

A

Tmxx

r

IB

r

IB

rBIB

o

)1026.1(104

constantty permeabili vacuum

2

2

2

1

67

o

o

internal

m

m

m

The magnetic field, B, is directly proportional

to the current, I, and inversely proportional

to the circumference.

ExampleA long, straight wire carries a current of 5.00 A. At one instant, a

proton, 4 mm from the wire travels at 1500 m/s parallel to the wire and in the same direction as the current. Find the magnitude and direction of the magnetic force acting on the proton due to the field caused by the current carrying wire.

))(1500)(106.1(

)004.0)(14.3(2

)5)(1026.1(

2

19

6

wireB

IN

oINEXB

BxF

xB

r

IBqvBF

m

5A

4mm+

v

2.51 x 10- 4 T

6.02 x 10- 20 N

X X X

X X X

X X X

X X X

X X X

X X XB = +z

v = +y

F = -x

Example

Two very long wires, separated by a distance of 9 mm, each have current

running through them. The first wire has a current of 5 A directed to the North,

while the second wire has a current of 7 A directed to the South.

a) Are the wires attracting or repelling?

b) Determine the magnitude of the force per unit length exerted on each

wire.

Sources of Magnetic Fields

In the last section, we learned that if a charged

particle is moving and then placed in an

EXTERNAL magnetic field, it will be acted upon by

a magnetic force. The same is true for a current

carrying wire.

The reason the wire and/or particle was moved

was because there was an INTERNAL magnetic

field acting around it. It is the interaction between

these 2 fields which cause the force.

Can we define this INTERNAL

magnetic field mathematically?

Magnetic Field of a moving charged particleThe magnetic field surrounding a moving charge can be

determined by the equation

Here we see that the FIELD is directly related to the CHARGE and

inversely related to the square of the displacement. The key in the case of

the B-Field is that particle MUST be moving and the vectors MUST be

perpendicular.

2

0sin

4 r

vqB

m

Magnetic Field of a current element• Begin by determining the magnetic field surrounding a

small current element in a current-carrying wire.

• Where r̂ (“r-hat”) is a unit vector, indicating that dl and r

must be perpendicular.

• The integral of this equation can be used to determine the

net magnetic field of an entire length of wire.

3

0ˆ

4 r

rldIBd

m

3

0

2

0

2

0

2

0

ˆ

4

sin

4

sin

4

sin

4

r

rldIBd

r

Idl

dtr

dqdldB

dt

dqI

dt

dlv

r

dqvdB

m

m

m

m

Biot-Savart Law

• Used to determine the magnetic field at a point in space near

a current-carrying wire.

• The equation may be rewritten as

• This is the “useful” form

3

0ˆ

4 r

rldIB

m

msin

4 2

0

r

dlIB

dl

I

B=?

Biot-Savart Law (wires)

m

m

m

m

sin4

dB

,ˆ4

B

sinr̂constant,4

,ˆ4

dB

2

0

3

0

0

3

0

r

Idl

dt

dqIr

dtr

dqdld

dt

dlv

rrr

dqv

This is for a current carrying element. The “dl” could represent a small

amount of a wire. To find the ENTIRE magnetic field magnitude at a point

away from the wire we would need to integrate over the length.

dl

I

B=?

Magnetic Field of a Straight WireConsider a wire carrying a current I. Determine the magnitude of the magnetic field

a distance R from the wire at any point.

210

00

0

0

2

2

0

2

0

2

0

2

0

sinsin4

coscos4

cos4

cos

4

cos4

sin4

cossinsin4

dB

21

m

m

m

m

m

m

m

IB

ddI

B

dI

rR

drIB

r

dyI

r

IdlB

r

Idl

θ1

α1

θ2

α2

dR

rdy

dR

Rrdy

dRdy

Ry

2

2

2

2sec

tan

R

ytan

Magnetic Field of a Straight WireConsider a straight wire of length L carrying a current I. Determine the magnitude of

the magnetic field a distance R from the wire at its center.

Magnetic Field of a Straight WireDetermine the magnitude of the magnetic field a distance R from a wire carrying

current I that is very long.

Magnetic Field of a Circular Current LoopDetermine the magnitude of the magnetic field at the center of a circular loop with

radius R carrying a current I.

Magnetic Field of a Circular Current LoopDetermine the magnitude of the magnetic field at a distance D along an axis

through the center of a circular loop with radius R carrying a current I.

α

α

Biot-Savart Law (wires)

What is the equation for the magnitude of the magnetic

field at the center of a current carrying loop?

Knowing this can be used

in conjunction with a

tangent galvanometer to

solve for the magnetic

field of Earth.

AP PHYSICS C

UNIT 10: MAGNETISM

PART 3

AMPERE’S LAW

Ampere’s Law

• Just as Gauss’s Law was an easier way to calculate the

electric field in symmetric situations, there is a similar

technique used for calculating magnetic fields.

• This is known as Ampere’s Law

• Where the integral is around a closed

path and I is the total current passing

through the area bounded by this path.

• A path is considered an Amperian Path when the magnetic

field, B, is constant at every point along the path.

IldB 0m

Ampere’s Law – Long Straight Wire

I

Bdl

When we SUM all of the

current carrying elements

around the PATH of a circle we

get the circle’s circumference.

Once again we see we get the

equation for the magnetic field

around a long straight wire.

This is the same as was derived from

the Biot-Savart Law

r

IB

IrB

IdlB

IldBenc

m

m

m

m

2

2

0

0

0

0

Since the path is Amperian and

B is constant, it can be

removed from the integral.

Caution

• Ampere’s Law only gives the magnetic field due to any

currents that cut through the area bounded by the integration

path.

• For the indicated integration path, the result of Ampere’s Law

is zero.

Example

Given three currents andtheir associated loops

For which loop is B • dlthe smallest?

A B C Same

The currents that are used in Ampere’s Law are the currents that are within the loop that is used for integration

Loop A has no current going through it whereas the other two loops do have currents going through them, the integral for Loop A is zero

A B C

Example

Given three currents and

their associated loops

Now compare loops B and C,

For which loop is B • dl

the greatest?

B C Same

The right hand side of Ampere’s law is the same for both loops,

therefore the integral for both loops is the same

A B C

Example – Long Cylindrical Conductor

• A long straight wire of radius R carries a current I that is

uniformly distributed over the circular cross section of the

wire. Find the magnetic field both outside the wire and inside

the wire.

R

riro

II

Example – Long Cylindrical ConductorA long straight wire of radius R carries a current I that is uniformly distributed

over the circular cross section of the wire. Find the magnetic field both

outside the wire and inside the wire.

Let’s look at the OUTSIDE field, ro > R

r

IB

IrB

IdlB

ooutside

o

enc

m

m

m

2

)2(

0

R

riro

II

Example – Long Cylindrical ConductorA long straight wire of radius R carries a current I that is uniformly distributed over the

circular cross section of the wire. Find the magnetic field both outside the wire and inside the wire.

Let’s look at the INSIDE field, ri < R

2

2

2

0

2

2

22

2

)2(

R

IrB

R

rIrB

IdlB

R

rII

R

I

r

I

oinside

o

enc

encenc

m

m

m

Since the current is

distributed throughout

the cross section we

can set up a ratio of

the currents as it

relates to the cross

sectional area.

We first need to identify exactly what is

the ENCLOSED current. It isn’t , “I”,

but rather a FRACTION of “I”.

R

riro

II

Example – Long Cylindrical ConductorHow could the magnetic field be graphically displayed?

rBR

IrB

rB

r

IB

oinside

ooutside

m

m

22

1

2

riro

R

riro

II

Fields produced by two parallel long wires

Two long wires are placed parallel to one another a distance d apart and each carry

a current I in the opposite direction, as shown below.

a) Do the wires attract or repel one another?• Repel. The fields are in the same direction.

b) Determine an expression for the magnitude of the

magnetic field midway between the wires.• The field around each wire is determined by B = μ0I/2πr

c) What is the direction of the magnetic field

between the wires?• Into the page

d) How much force per unit length is exerted on the left wire

by the right wire?

I1 I2

d

d

II

L

F

d

II

L

F

BIL

F

LBIF

m

m

2

2

sin

2101

20

1

1

21

1

211

190sin

90

o

d

IB

m

2

20

2

Example

A current I flows in an infinite straight wire in the

+z direction as shown. A concentric infinite cylinder

of radius R carries current 2I in the -z direction.

What is the magnetic field Bx(a) at point a, just

outside the cylinder as shown?

(a) Bx(a) < 0 (b) Bx(a) = 0 (c) Bx(a) > 0

The loop used for application of Ampere’s Law will be outside the cylinder

The total current going through this loop is the sum of the two current

yielding a current of –I. This gives a magnetic that points in the positive x

direction at a

xx

xx

xx

x

x

2II

a

b

x

y

Example

A current I flows in an infinite straight wire in the

+z direction as shown. A concentric infinite cylinder

of radius R carries current 2I in the -z direction.

What is the magnetic field Bx(b) at point b, just

inside the cylinder as shown?

(a) Bx(b) < 0 (b) Bx(b) = 0 (c) Bx(b) > 0

• This time, the Ampere loop only encloses current I which is in the +z

direction — the loop is inside the cylinder!

• The current in the cylindrical shell does not contribute to at point b.

xx

xx

xx

x

x

2II

a

b

x

y

Applications of Ampere’s Law – A Solenoid

A solenoid is basically a bunch of loops of

wire that are tightly wound. It is analogous to

a capacitor which can produce a strong

electric field. In this case it can produce a

strong MAGNETIC FIELD.

Solenoids are important in engineering as

they can convert electromagnetic energy

into linear motion. All automobiles use

what is called a “starter solenoid”. Inside

this starter is a piston which is pushed out

after receiving a small amount of current

from the car’s battery. This piston then

completes a circuit between the car’s

battery and starter motor allowing the car

to operate.

Applications of Ampere’s Law – A

SolenoidDetermine the magnetic field at the center of a solenoid.

The first thing you must understand is what is the

enclosed current. It is basically the current, I, times the #

of turns you enclose, N.

When you integrate all of the small current elements

they ADD up to the length of the solenoid, L

nIB

L

NnI

L

NB

NILB

IdlB

osolenoid

enc

m

m

m

m

length per turns# ,

)()(

0

0

0

It is important to understand that when you

enclose a certain amount of turns that the

magnetic field runs through the center of

the solenoid. As a result the field lines and

the length of the solenoid are parallel. This

is a requirement for Ampere’s Law.

Example

A solenoid has a length L =1.23 m and an inner diameter d =3.55

cm, and it carries a current of 5.57 A. It consists of 5 close-

packed layers, each with 850 turns along length L. What is the

magnetic field at the center?

B

xB

nIB

L

Nn

osolenoid

)57.5(23.1

8505)1026.1(

length per turns#

6

m

0.024 T