Part B: Electromagnetism

Chapter 4: Magnetism (Chapter 28, 29 in textbook)

4.1. The Magnetic Field 4.2. The Hall Effect 4.3. Motion of a Charged Particle in a Magnetic Field 4.4. Magnetic Force on a Current-Carrying Wire 4.5. Torque on a Current-Carrying Coil 4.6. The Magnetic Dipole Moment 4.7. The Biot–Savart Law 4.8. Ampere’s Law 4.9. The Magnetic Field of a Solenoid and a Toroid 4.10. The Magnetic Field of a Current-Carrying Coil

Overview In this part, we study the science of magnetic fields, including what produces a magnetic field, how a magnetic field can produce a magnetic force on a moving charged particle. The applications of magnetic fields are countless, e.g., audiotape, videotape, CD, DVD players, TVs, computers, telephones, medical devices,…

4.1. The Magnetic Field 4.1.1. What produces a Magnetic Field? •We have discussed how a charged plastic rod produces a vector field - the electric field at all points in the space around the rod. •Here, we have a magnet that produces a vector field – the magnetic field at all points in the space around the magnet Examples: •Permanent magnets: magnets at the door of refrigerators •Electromagnets: a wire core coil is wound around an iron core and a current is sent through the coil. The magnetic field is produced by the flow of electric current, and the magnetic field disappears when the current is turned off

B

We have studied that an electric charge sets up an electric field that affects other electric charges. Here, we might expect that a magnetic charge sets up a magnetic field that can then affect other magnetic charges. Such a magnetic charge is called a magnetic monopole (a new elementary particle?) predicted by certain theories but its existence has NOT been confirmed. The key question is “How then are magnetic fields set up?” There are two ways: (1) Moving electrically charged particles, e.g., a current in a wire,

creating magnetic fields (2) Elementary particles such as protons, electrons, have an

intrinsic magnetic field around them. This field is a basic characteristic of the particles such as their mass, their electric charge. In some materials, the magnetic fields of electrons add together to give a net magnetic field around the material (magnets)

4.1.2. The Definition of : As the existence of the magnetic charge has not been confirmed, so we can not use the way of definition of electric field Here. We must use another way to define : • firing a charged particle through the point at which B is to

be defined with various directions and speeds • determining the force acting on the particle at that point • After many such trials, we can find when the particle’s

velocity is along a particular axis, is zero. For all other directions, FB ~ vsinφ, where φ is the angle between the zero-force axis and .And the direction of is always perpendicular to

B

qFE

=

B

BF

BF

v

v

BF

vqFB B

||=

We can summarize all the results above by: φ is the angle between the directions of and c. The Magnetic Force Acting on a Particle: We use the right-hand rule to determine the direction: the fingers (of the right-hand) sweep into through the smaller angle φ, the thumb points in the direction of , then we consider the sign of charge q Two special cases: • FB = 0 if φ = 00 or φ = 1800

• FB is maximum if φ = 900

BvqFB

×=

φsin|| vBqFB =

v

B

v

B

Bv

×

BF

• If q > 0 (figure b): the force is directed along the thumb • If q < 0 (figure c): the force is directed opposite the thumb

The tracks of two electrons (e-) and a positron (e+) in a bubble chamber in a uniform magnetic field, which is

directed out of the plane of the page

γ ray

v

B

Bv

×

• The SI unit for B is called the tesla (T):

nd)meter/seco(coulomb)(newton1 T 1 tesla 1 ==

A.mN

er)econd)(met(coulomb/snewton1 T 1 1==

• An earlier non-SI unit for B is the gauss (G):

gauss 10 tesla 1 4=

4.1.3. Magnetic Field Lines: Magnetic fields can be represented by field lines with the following rule: (a) the direction of the tangent to a magnetic field line at any point gives the direction of B at that point (b) the spacing of the lines represents the magnitude of B, it means the magnetic field is stronger where the lines are closer together, and conversely (see Figure a for a bar magnet). Note that: •All the lines pass through the magnet and they all form closed loops. •A magnet has two poles (magnetic dipole): north pole and south pole. The field lines are from the north pole to the south pole •Opposite magnetic poles attract each other, and like magnetic poles repel each other

A horseshoe magnet

A C-shaped magnet

4.1.4. Thomson’s Experimental Apparatus: Discovery of the Electron

Crossed fields: An electric field E and a magnetic field B can produce a force on a charged particle, when they are perpendicular to each other, they are crossed fields.

A modern version of Thomson’s apparatus for measuring the ratio of mass to charge for the electron (considered to be the “discovery of the electron”)

In this arrangement, electrons from the hot filament are forced up by electric field E and down by magnetic field B, so the forces are in opposition Thomson carried out the following steps: 1. Set E = 0 and B = 0, no deflection of the electron beam 2. Turn on E and measure the beam deflection (see Sample

problem, p.593) L: length of the plates 3. Maintaining E, turn on B and adjust its value until the beam

returns to the undeflected position, FE = FB

So: (1) & (2):

)1(2

||2

2

mv

ELqy =

vBqvBqEq ||)90sin(|||| 0 ==

)2(BEv =

yELB

qm

2||

22=

4.2. The Hall Effect: • The effect was discovered by Edwin H. Hall in 1879 when he was a graduate student at the Johns Hopkins University • The Hall effect mentions that conduction electrons in a wire are deflected by a magnetic field

(a) Electrons in a copper strip are deflected by B, moving to the right edge of the strip

(b) The separation of positive and negative charges produces E as shown, resulting in FE to the left

(c) An equilibrium is established as FB=FE: Using a voltmeter, we can measure V and determine which edge is at higher potential to check the electron deflection.

EdV =

http://hyperphysics.phy-astr.gsu.edu/

l

• The Hall effect also allows us to find out whether the charge carriers in a conductor are positive or negative:

negative charge positive charge

For an equilibrium: Drift speed: We can derive the number density: where l = A/d is the thickness of the strip

BeveE d=

neAi

neJvd ==

VleBin =

4.3. Motion of a Charged Particle in a Magnetic Field: • We consider a beam of electrons moving in a region of uniform magnetic field B as shown. The magnetic force acts on an electron, causing the electrons moving along a circular path:

rvmqvBFB

2== • The radius of the circular path:

qBmvr =

• the period:

qBm

vrT ππ 22==

• the angular frequency:

mqBf == πω 2

Helical Paths: In a general case, if the velocity of the charge has a component parallel to the magnetic field, the charge will move in a helical path about the B field direction: The V|| component determines the pitch p of the helix

φφ sincos|| vvvv == ⊥ and

uniform B

nonuniform B – magnetic bottle

The auroral oval surrounding Earth’s geomagnetic north pole

movie

movie

Cyclotrons and Synchrotrons: They are particle accelerators that can give the particles enough kinetic energy to slam into a solid target. Then we analyze the debris of the collisions to study the subatomic particles of matter • The Cyclotron: A compact type of particle accelerators, in the cyclotron the particles circulate and they are accelerated to ~10 MeV for a proton

The Proton Synchrotron: It is a particular type of cyclic particle accelerator, originally from the cyclotron. The protons can be accelerated to ~ 1 TeV (1012 eV)

The Grenoble synchrotron

The Large Hadron Collider (LHC) is the world’s largest and highest- energy particle accelerator

4.4. Magnetic Force on a Current-Carrying Wire: Since magnetic fields exert a force on electrons moving in a wire, so the force is then transmitted to the wire itself. Consider a length L of the wire, all the electrons will drift past plane xx in a time: The charge pass through the plane:

dvLt /=

dvLiitq ==

Force acting on the wire of length L:

iLBBvviLBqvF dd

dB === φsin

In general, force acting on a length L of straight line: is a length vector with magnitude L and direction along the wire segment in the direction of the current Note: If the field is not uniform or the wire is not straight, we can use the following equation for small segments:

BLiFB

×=

φsiniLBFB =

L

BLidFd B

×=

4.5. Torque on a Current-Carrying Coil: In this section, we study the magnetic forces exerting on a current-carrying loop, a basic element of electric motors. Consider a single current-carrying loop immersed in a B field as shown: Two magnetic forces F and –F produce a torque on the loop To define the orientation of the loop in the B field, we use a normal vector (curl the fingers of your right hand in the current direction, your thumb points in the direction). F2 and F4 cancel out each other: their net force is zero and their net torque is also zero

n

n

Animation

Animation

θ θ n

1F

3F

B

2b

2b

a

b

The torque due to forces F1 and F3:

θτ

θθτ

sin'

)sin2

()sin2

('

iabB

biaBbiaB

=

+=

θτ sin' iAB=where A = ab: the area of the loop Torque τ’ tends to rotate the loop to align its normal vector with the direction of

Torque on a current-carrying coil: For a coil of N loops, the total torque acts on the coil:

θττ sin' NiABN ==

B

moment arm

4.6. The Magnetic Dipole Moment: A current-carrying coil in a magnetic field acts like a bar magnet, so the coil is considered to be a magnetic dipole. Thus, we can define a magnetic dipole moment of the coil: : its direction is the direction of the fingers of your right hand curl around the coil in the current direction, your extended thumb points the direction N: the number of turns in the coil i: the current A: the area enclosed by each turn Unit: A.m2

θ: the angle between and

µ

NiA=µ

θµτ sinB=

µ

n

µ

B B

×= µτ

µ

Orientation Energy of a Magnetic Dipole: A magnetic dipole in an external magnetic field has a magnetic potential energy (unit: joule): • θ = 00, lowest energy:

• θ = 1800, highest energy:

Work done on the dipole by the magnetic field: Work done on the dipole by the applied torque:

θµµθ cos)( BBU −=−=

BU µθ −=)(

BU µθ +=)(

)( if UUUW −−=∆−=

ifa UUWW −=−=

Checkpoint: Rank the orientations of a magnetic dipole moment in a magnetic field according to (a) the magnitude of the torque on the dipole and (b) the potential energy of the dipole, greatest first

(a)all tie

(b) 1-4, 2-3

),sin( BB

µµτ =

),cos()( BBU

µµθ −=

Homework: 3, 5, 9, 13, 21, 25, 29, 31, 40, 45, 49, 51, 57,

62 (pages 757-761)

Overview In this lecture, we study magnetic fields produced by a moving charged particle or a current. This feature of electromagnetism is the combination of electric and magnetic effects and it has become extremely important because of its application in our life

4.7. The Biot–Savart Law (Calculating the Magnetic Field Due to a Current): Problem: We need to calculate the magnetic field B at point P due to a current Method: We use the same principle as used to calculate the electric field due to a charge distribution: • We divide (mentally) the wire into differential elements • A differential current-length element • From experiments, the field where : is the angle between and : is the permeability constant

sd

sid

Bd

20 sin

4 r

idsdB θπµ

=

θ sd r

0µ

AmTAmT /.1026.1/.104 670

−− ×≈×= πµ

The Biot-Savart law: We will use the law above to calculate the magnetic field at a point due to various distributions of current 4.7.1. Magnetic Field Due to a Current in a Long Straight Wire: we calculate the B field due to the upper half of the wire as shown, the total B is:

30

4 r

rsidBd ×

=πµ

20 sin

4 r

idsdB θπµ

=

∫∫∞∞

==

02

0

0

sin2

2r

dsidBB θπ

µ

We also have: (see Appendix E for the integral) To determine the B direction, we use the right hand rule: Grasp the element in your right hand with your extended thumb pointing in the direction of the current. Your fingers will then naturally curl around in the direction of the magnetic field lines due to that element

22 Rsr +=

22)sin(sin

Rs

R

+=−= θπθ

∫∞

+=

02/322

0)(2 Rs

RdsiBπ

µ

Ri

Bπµ2

0=

Iron filings that have been sprinkled onto cardboard collect in concentric circles when current is sent through the central wire

4.7.1. Magnetic Field Due to a Current in a Circular Arc of Wire: Problem: Find the magnetic field produced at a point by a current in a circular arc of wire Method: We calculate the field produced by a single current-length element, then integrate to find the net field by all the elements Example: An arc-shaped wire with angle φ, radius R, current i with θ = 900: with ds = Rdφ:

20

20

4sin

4 R

ids

r

idsdBπµθ

πµ

==

∫∫∫ ===φφ

φπµφ

πµ

0

0

02

044

dRi

R

iRddBB

• So, the magnitude of the field produced by a circular arc of wire: • For a full circle, the field at the center: Note: To determine the direction of the magnetic field, we use the right-hand rule

Ri

Bπφµ

40=

Ri

RiB

242 00 µππµ==

Checkpoint: The figure shows three circuits consisting of straight radial lengths and concentric circular arcs (either half- or quarter-circles of radii r, 2r, and 3r). The circuits carry the same current. Rank them according to the magnitude of the magnetic field produced at the center of curvature (the dot), greatest first (a) (b) (c) Rank: a, c, b

Ri

Bπφµ

40=

ri

ri

ri

BBB3434000

21µ

ππµ

ππµ

=+=+=

ri

ri

riBBB

6344000

12µ

ππµ

ππµ

=−=−=

ri

ri

ri

riB

4813

24224340000 µπ

πµπ

πµ

ππµ

=++=

321 BBBB ++=

1

2

1 2

1

2 3

Magnetic Field Due to Brain Activity

The magnetic fields detected in MEG (magneto-encephalo-graphy), a procedure to monitor the human brain, are probably produced by pulses (e.g., when reading) along the walls of the fissures (crevices) on the brain surface: to measure such a very small field, we need to use an instrument called SQUIDs (Superconducting Quantum Interference Devices)

Tm

mAAmTdB 12022

367105.290sin

)102(

)101)(1010(4

)/.104( −−

−−−×=

×

×××=

ππ

• In a typical pulse: i = 10 µA • The conducting path length: 1 mm • Point P at r = 2cm • Angle θ = 900

MEG (magnetoencephalography) Electroencephalography

4.7.2. Force Between Two Parallel Currents: Problem: Two long parallel wires carrying currents exert forces on each other. Calculate those forces The field Ba produced by current a at the site of wire b: So, the force Fba acting on b from a: The magnitude: • Applying the same steps, we can also calculate Bb at current a and Fab acting on current a is shown in the figure these forces pull the currents close to each other • If the two currents are antiparallel, the forces push the currents apart

diB a

a πµ20=

abba BLiF

×=d

iLiLBiF ba

abba πµ

290sin 00 ==

bB

abF

Rail Gun: An application is based on the B field produced by two antiparallel currents: • Conducting fuse (e.g., a piece of copper) will melt and vaporize after the current passes through it, creating a conducting gas • The magnetic field produced by the two currents are directed downward between the rails • The field exerts a force on the gas, which is forces outward along the rails and thus the gas pushes the projectile Example: A projectile of 5 tonnes (5000 kg) can be accelerated to a speed of 10 km/s within 1 ms

Checkpoint: The figure shows three long, straight, parallel, equally spaced wires with identical currents either into or out of the page. Rank the wires according to the magnitude of the force on each due to the currents in the other two wires, greatest first. The net force acting on curent b: For a: For c:

cB

aB

baF

bcF

2222

2000

dLi

diLi

diLi

FFF bcbabcbab π

µπ

µπ

µ=+=+=

bB

cB

abF

acF

21

2222

2000

dLi

diLi

diLiFFF bcba

acaba πµ

πµ

πµ

=−=−=

aB

bB

cbF

caF

23

2222

2000

dLi

diLi

diLiFFF bcba

cacbc πµ

πµ

πµ

=+=+=

4.8. Ampere’s Law: • In the previous chapter, we use Gauss’ law to determine the net electric field due to some symmetric distributions of charges (planar, cylindrical, spherical symmetry) • Here, to determine the net magnetic field due to some symmetric distributions of currents we use Ampere’s law:

where ienc is the net current encircled in the Amperian loop

∫ = encisdB 0µ

• To apply Ampere’s law: o Arbitrarily choose the direction of integration o Arbitrarily assume to be generally in the direction of integration o Use the curled-straight right-hand rule (see the next slide) to assign a plus sign or a minus sign to each of the currents

B

The curled-straight right hand rule: Curl your right hand around the Amperian loop, with the

fingers pointing in the direction of integration. A current through the loop in the general direction of your outstretched thumb is assigned a plus sign, and a current generally in the opposite direction is assigned a minus sign Question: Why current i3 contributes

to the B field magnitude on the left side of the equation above but i3 is not in the right side? Because its contributions to the B field cancel out as the integration is made around a closed loop. In contrast, the contributions of an encircled current to the magnetic field do not cancel out

21 iii −=enc

)(cos 2100 iiidsB −==∫ µµθ enc

4.8.1. The Magnetic Field Outside a Long Straight Wire with Current: Here, we use Ampere’s law to find the B field at a point outside and produced by a long, straight wire. The direction of integration is counterclockwise: Using the right-hand rule, the current i is positive: so, we obtain: This is the field that we derived using the Biot-Savart law, but here the calculation is quite simple

∫ ∫ == )2(cos rBdsBdsB πθ

irB 0)2( µπ =

riB

πµ2

0=

4.8.2. The Magnetic Field Inside a Long Straight Wire with Current: We use Ampere’s law to find the B field inside a long, straight wire with current. The direction of integration is counterclockwise: If the current is uniformly distributed: so, we obtain: So, B = 0 at the center and maximum at the surface

∫ ∫ == )2(cos rBdsBdsB πθ

2

2

R

riiπ

π=enc

⇒= 2

20)2(

R

rirBπ

πµπ rR

iB

= 2

02π

µ

4.9. The Magnetic Field of a Solenoid and a Toroid: 4.9.1. Magnetic Field of a Solenoid: Solenoid: A long, tightly wound helical coil of wire The magnetic field produced by a solenoid carrying current i is shown: the field is strong and uniform at interior points but relatively weak at external points

A vertical cross section through the central axis of a current-carrying solenoid

The field direction along the solenoid axis is determined by the curled-straight right-hand rule

Calculate the magnitude of the magnetic field inside an ideal solenoid: Ampere’s law: For external points of an ideal solenoid: B = 0, so the third integral c d is zero where n is the number of turns per unit length

encisdB 0µ∫ =

∫∫

∫∫∫

++

++=

a

d

d

c

c

b

b

a

sdBsdB

sdBsdBsdB

BhBh =+++= 000

)(nhiienc =

inB 0µ=

4.9.2. Magnetic Field of a Toroid: Toroid: It is a solenoid curved until its two ends meet The field inside a toroid can be calculated using Ampere’s law. We choose an Amperian loop as shown and the direction of integration is clockwise: N: the total number of turns So, the field in a toroid is not uniform like a solenoid, B ~ 1/r The direction of magnetic fields is also given by the curled-straight right-hand rule

iNirB enc 002 µµπ ==

riNB 1

20π

µ=⇒

4.10. The Magnetic Field of a Current-Carrying Coil:

Recall: A current-carrying coil behaves as a magnetic dipole, if we place it in an external magnetic field Bext, a torque acting on it: N: the number of turns of the coil Magnetic field of a Coil: The field at a point on the central axis of the coil: if z >> R:

B

×= µτNiA=µ

)1()(2

)( 2/322

20

zR

iRzB

+=

µ

3

202

)(z

iRzB

µ≈

For a coil of N turns and A is the area of the loop: and have the same direction So, we have 2 ways to consider a current-carrying coil as a magnetic dipole: (1) it experiences a torque in an external magnetic field (2)it generates its own intrinsic magnetic field, acting

as a magnet

30

2)(

z

NiAzBπ

µ=

B µ

30

2)(

zzB µ

πµ

=

Homework:

4, 7, 12, 16, 18, 22, 35, 38, 43, 46, 49, 50, 57, 62 (pages 783-788)