Physics for Scientists & Engineers 2unitenphyf115.weebly.com/uploads/7/4/4/8/74480909/chapter_10.pdf · June 1, 2016 University Physics, Chapter 27 2 Permanent Magnets Examples of

June 1, 2016 1

Magnetism

Permanent magnets

Earth’s magnetic field

Magnetic force

Motion of charged particles in magnetic fields

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University Physics, Chapter 27

June 1, 2016 University Physics, Chapter 27

2

Permanent Magnets

Examples of permanent magnets include refrigerator magnets and magnetic door latches

They are all made of compounds of iron, nickel, or cobalt

If you touch an iron needle to a piece of magnetic lodestone, the iron needle will be magnetized

If you then float this iron needle in water, the needle will point toward the north pole of the Earth (approximately!)

We call the end of the magnet that points north the north pole of the magnet and the other end the south pole of the magnet


3

Permanent Magnets - Poles

Magnets exert forces on one another --- attractive or repulsive depending on orientation.

If we bring together two permanent magnets such that the two north poles are together or two south poles are together, the magnets will repel each other

If we bring together a north pole and a south pole, the magnets will attract each other


4

Magnetic Field Lines (1)

Permanent magnets interact with each other at a distance, without touching

In analogy with the electric field, we define a magnetic field to describe the magnetic force

As we did for the electric field, we may represent the magnetic field using magnetic field lines

The magnetic field direction is always tangent to the magnetic field lines


5

Magnetic Field Lines (2)

The magnetic field lines from a permanent bar magnet are shown below

Two dimensional computer calculation Three dimensional real-life


6

Broken Permanent Magnet

If we break a permanent magnet in half, we do not get a separate north pole and south pole

When we break a bar magnet in half, we always get two new magnets, each with its own north and south pole

Unlike electric charge that exists as positive (proton) and negative (electron) separately, there are no separate magnetic monopoles (an isolated north pole or an isolated south pole)

Scientists have carried out extensive searches for magnetic monopoles; all results are negative

Magnetism is not caused by magnetic particles! Magnetism is caused by electric currents

June 1, 2016 University Physics, Chapter 27 7

Magnetic Field Lines

For the electric field, the electric force points in the same direction as the electric field and the electric force was defined in terms of a positive test particle

However, because there is no magnetic monopole, we must employ other means to define the magnetic force

We can define the direction of the magnetic field in terms of the direction a compass needle would point

A compass needle, with a north pole and a south pole, will orient itself in equilibrium such that its north pole points in the direction of the magnetic field

Thus the direction of the field can be measured at any point by moving a compass needle around in a magnetic field and noting the direction that the compass needle points


8

Magnetic Force

We define the magnetic field in terms of its effect on an electrically charged particle (q).

Recall that an electric field exerts a force on a particle with charge q given by

A magnetic field exerts no force on a stationary charge

But a magnetic field does exert a force on a charge that moves across the field

The direction of the force is perpendicular to both the velocity of the moving charged particle and the magnetic field; the magnetic force is sideways

The Lorentz force

( )F qE x

( )F qv B x


9

Right Hand Rule (1)

The direction of the cross product (v x B ) is given by the right hand rule

To apply the right hand rule

• Use your right hand!

• Align thumb in the direction of v

• Align your index finger with the magnetic field

• Your middle finger will point in the direction of the cross product v x B


Right Hand Rule (2)

What about the sign of the charge?

• If q is positive, then F is in the same direction as v x B

• If q is negative, F is in the opposite direction


Magnitude of Magnetic Force

The magnitude of the magnetic force on a moving charge is

… where is the angle between the velocity of the charged particle and the magnetic field.

Do you see that there is no magnetic force on a charged particle moving parallel to the magnetic field? (Because is zero and sin(0)=0)

Do you see when the magnetic force is most strong? (Max. force is for = 90 degrees; then F = qvB)

sinBF qvB


12

Units of Magnetic Field Strength

The magnetic field strength has received its own named unit, the tesla (T), named in honor of the physicist and inventor Nikola Tesla (1856-1943)

A tesla is a rather large unit of the magnetic field strength

Sometimes you will find magnetic field strength stated in units of gauss (G), (not an official SI unit)

1 T = 1 Ns

Cm 1

N

Am

-41 G = 10 T 10 kG = 1 T

Check unit consistency: F = q v B N = C (m/s) T


13

Example: Proton in B Field

Consider a region of uniform magnetic field (green dots) coming out of the page with magnitude B = 1.2 mT. A proton with kinetic energy K = 8.4810-13

J enters the field, moving vertically from the bottom to the top.

Question: Calculate the magnetic deflecting force on the proton. Answer: Use F = qvBsin. First, figure out the velocity v.

K 1

2mv2 v

2K

m

v 2 8.48 1013 J 1.67 1027 kg

3.2 107 m/s


Direction of F: Use Right Hand Rule

Example: Proton in B Field (2)

Small force but large acceleration for light particle

19 7 3

15

1512 2

27

sin

1.60 10 C 3.2 10 m/s 1.2 10 T sin 90

6.1 10 N

6.1 10 N3.7 10 m/s

1.67 10 kg

F qvB

F

F

Fa

m

Angle between B-field direction and velocity of the proton: deg


15

Example: Cathode Ray Tube (1)

The mass of an electron is 9.109410-31 kg while the elementary charge is 1.602210-19 C.

Question: Calculate the velocity of the electrons in the beam after leaving the electron gun.

Answer: 2 61

2 implies 6.92 10 m/s

K U qV

mv eV v

Consider a cathode-ray tube.

In this tube electrons form an electron beam when accelerated horizontally by a voltage of 136 V in an electron gun.


16

Example: Cathode Ray Tube (2)

Question: If the tube is placed in a uniform magnetic field, in what direction is the electron beam deflected?

Answer:

Question: Calculate the magnitude of acceleration of an electron if the field strength is 3.65×10-4T.

Answer:

F ma qvB

downward

1.6022 1019 C 6.92 106 m/s 3.65104 T

9.1094 1031 kg 4.44 1014 m/s2

qvBa

m


17

Particle Orbits in a Uniform B (1)

Tie a string to a rock and twirl it at constant speed in a circle over your head.

The tension of the string provides the centripetal force that keeps the rock moving in a circle.

The tension on the string always points to the center of the circle and creates a centripetal acceleration.

A particle with charge q and mass m moves with velocity v perpendicular to a uniform magnetic field B.

The particle will move in a circle with a constant speed v and the magnetic force F = qvB will keep the particle moving in a circle.


Particle Orbits in Uniform B (2)

Recall centripetal acceleration

Newton’s second law

So, for a charged particle q in circular motion in a magnetic field B

Or

m

v2

r qvB

m

v

r qB

p

r qB

a

v2

r

F ma


19

Example: Moving Electrons

In this photo, an electron beam is initially accelerated by an electric field.

Then the electrons move in a circle perpendicular to the constant magnetic field created by a pair of Helmholtz coils.

v

F

Question: Is the magnetic field into the page or out of the page? (Remember that the magnetic force on an electron is opposite that on a proton.) Answer: The magnetic field is out of the page.


Orbits in a Constant Magnetic Field

If a particle performs a complete circular orbit inside a constant magnetic field, then the period of revolution of the particle is just the circumference of the circle divided by the speed

From the period we can get the frequency and angular frequency

The frequency of the rotation is independent of the speed of the particle.

Isochronous orbits Basis for the cyclotron

2 2r mT

v qB

1

2

qBf

T m 2

qBf

m


21

Cyclotrons A cyclotron is a particle accelerator

The D-shaped pieces (descriptively called “dees”) have alternating electric potentials applied to them such that a positively charged particle always sees a negatively charged dee ahead when it emerges from under the previous dee, which is now positively charged

The resulting electric field accelerates the particle

Because the cyclotron sits in a strong magnetic field, the trajectory is curved

The radius of the trajectory is proportional to the momentum, so the accelerated particle spirals outward


22

Force on a Current Carrying Wire (1)

Consider a long, straight wire carrying a current i in a constant

magnetic field B

The magnetic field will exert a force on the moving charges in the wire

The charge q flowing in the wire in a given time t in a length L of wire is given by

where v is the drift velocity of the electrons

Lq ti i

v


23

Force on a Current Carrying Wire (2)

The magnitude of the magnetic force is then

is the angle between the current and the magnetic field

The direction of the force is perpendicular to both the current and the magnetic field and is given by the right hand rule

This equation can be expressed as a vector cross product

iL represents the current in a length L of wire

for F iLBF iL B L B

F qvBsin L

vi

vBsin iLBsin


24

The force on a current carrying wire is

The magnitude of the magnetic force is

iL represents the current in a length L of wire is the angle between the current and the

magnetic field The direction of the force is perpendicular to both

the current and the magnetic field and is given by the right hand rule

Reminder: Force on a Current Carrying Wire

sinF iLB

F iL B


25

Parallel Current Carrying Wires (1)

Consider the case in which two parallel wires are carrying currents

The two wires will exert a magnetic force on each other because the magnetic field of one wire will exert a force on the moving charges in the second wire

The magnitude of the magnetic field created by a current carrying wire is given by

This magnetic field is always perpendicular to the wire with a direction given by the right hand rule

0( )2

iB r

r


26


Let’s start with wire one carrying a current i1 to the right

The magnitude of the magnetic field a distance d from wire one is

Now consider wire two carrying a current i2 in the same direction as i1 placed a distance d from wire one

The magnetic field due to wire one will exert a magnetic force on the moving charges in the current flowing in wire two

0 11

2

iB

d

The charge q2 flowing in wire two in a given time t in a length L of wire is given by

where v is the drift speed of the charge carriers

The magnetic force is then

Putting in our expression for B1 we get



2 2 2

Lq ti i

v

2 1 2 1

LF qvB i vB i LB

v

0 1 0 1 212 2

2 2

i i i LF i L

d d


28

The figure shows three long, straight, parallel, equally spaced wires with identical currents either into or out of the page.

Question: What is the force on wire (a) due to the other two?

Answer: Step 1: Find the net magnetic field from wires (b) and (c): Bbc

Step 2: Find the force on wire (a)

Bb at (a) is down

Bc at (a) is up

The net field is

Example: Straight Parallel Wires

Force on (a) by (b) and (c)

Bb 0ib

2d Bc

0ic

2 2d

Bbc 0i

2d0i

4d

0i

4d

abc a bcF i L B

Fabc iaLBbc 0i

2L

4d

Electric motors rely on the magnetic force exerted on a current carrying wire

This force is used to create a torque that turns a shaft A simple electric motor is depicted below consisting of a

single loop carrying current i in a constant magnetic field B

The two magnetic forces, F and -F, shown in the figure are of equal magnitude and opposite direction

These forces create a torque that tends to rotate the loop around its axis


Torque on a Current-Carrying Loop (1)



As the coil turns in the field, the forces on the sides of the loop perpendicular to the magnetic field will change

The forces act on the sides of the square loop below, where is the angle between a normal vector, n, and the magnetic field B

The normal vector is perpendicular to the plane of the wire loop and points in a direction given by the right hand rule based on the current flowing in the loop


31


Here the current is flowing upward in the top segment and downward in the lower segment as illustrated by the arrow feathers and arrowhead

The force of each of the vertical segments is

The force on the other two sides is parallel or anti-parallel

to the axis of rotation and cannot cause a torque

(remember )F BF iaB iL


32


The sum of the torque on the upper side plus the torque on the lower side gives the torque exerted on the coil about the center of the loop

where A = a2

1 iaB

a

2

sin iaB

a

2

sin ia2Bsin iABsin

r F


33

Example: Current Carrying Loop (1)

A single-turn current loop, carrying a current of 4 A, is in the shape of a triangle with sides 50, 120, and 130 cm. The loop is in a uniform magnetic field of strength B = 75 mT (direction parallel to the 130 cm side of the loop).

Question: What is the magnitude of the magnetic force on the 130 cm side?

Answer:

For the 130 cm side:

0

F iL B

iL B F


Example: Current Carrying Loop (2)

Question: What is the magnitude of the magnetic force on the 120 cm side?

Answer: Look at the x and y components of the magnetic field!

1

120

120

cos

sin

tan 50 cm /120 cm 22.6

There is no contribution from on since

Force due to is

4 A 1.20 m 0.075 T sin 22.6 0.138 N

x

y

x x x

y y

x y

B B

B B

B iL B iL

B iL B

F iL B


Reminder - Cross Product

F qv B

F iL B

Force F exerted on

charge q with velocity v

moving at angle relative

to magnetic field B.


Magnetic Flux

To quantify the amount of magnetic field lines we define the magnetic flux in analogy to the electric flux

When we introduced Gauss’s Law for the electric field, we defined the electric flux as

For the magnetic field, we can define magnetic flux in analogy as

The unit of magnetic flux is the weber (Wb)

E E dA

B B dA

21 Wb 1 Tm


Consider the special case of a flat loop of area A in a constant magnetic field B

In this case we can re-write the magnetic flux as

• is the angle between the surface

normal vector of the plane of the loop and the magnetic field

If the magnetic field is perpendicular to the plane of the loop • = 0, B = BA

If the magnetic field is parallel to the plane of the loop • = 90, B = 0

Magnetic Flux - Special Case

cosB BA

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Physics for Scientists & Engineers 2unitenphyf115.weebly.com/uploads/7/4/4/8/74480909/chapter_10.pdf · June 1, 2016 University Physics, Chapter 27 2 Permanent Magnets Examples of