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Chapter 26: Magnetism: Force and Field. Magnets. Magnetism. Magnetic forces. Magnetism. Magnetic field of Earth. Magnetism. S. N. S. N. S. N. Magnetic monopoles?. Perhaps there exist magnetic charges, just like electric charges.

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magnetism2

S

N

S

N

S

N

  • Magnetic monopoles?

Perhaps there exist magnetic charges, just like electric charges.

Such an entity would be called a magnetic monopole (having + or magnetic charge).

How can you isolate this magnetic charge?

Try cutting a bar magnet in half:

Magnetism

Even an individual electron has a magnetic “dipole”!

  • Many searches for magnetic monopoles—the existence of which would explain (within framework of QM) the quantization of electric charge (argument of Dirac)
  • No monopoles have ever been found:
magnetism3

Orbits of electrons about nuclei

Intrinsic “spin” of electrons (more important effect)

  • Source of magnetic field
  • What is the source of magnetic fields, if not magnetic charge?
  • Answer: electric charge in motion!
    • e.g., current in wire surrounding cylinder (solenoid) produces very similar field to that of bar magnet.
  • Therefore, understanding source of field generated by bar magnet lies in understanding currents at atomic level within bulk matter.

Magnetism

magnetism6

Magnetic force (cont’d)

Components of the magnetic force

Magnetism

magnetism7

B

B

B

x x x x x x

x x x x x x

x x x x x x

® ® ® ® ®

® ® ® ® ®

­ ­ ­ ­ ­ ­ ­ ­

­ ­ ­ ­ ­ ­ ­ ­

v

v

v

´

q

q

q

F

F = 0

F

  • Magnetic force (cont’d)

Magnetic force

Magnetism

magnetism8

Magnetic force (cont’d)

Units of magnetic field

Magnetism

slide12

Magnetic Field Lines and Flux

  • Magnetic field lines
magnetic field lines and flux

Magnetic field lines

Magnetic Field Lines and Flux

slide14

Magnetic Field Lines and Flux

  • Magnetic field lines (cont’d)
slide15

Electric Field Linesof an Electric Dipole

Magnetic Field Lines of a bar magnet

Magnetic Field Lines and Flux

  • Magnetic field lines (cont’d)
slide16

Magnetic Field Lines and Flux

  • Magnetic field lines (cont’d)
slide17

Magnetic Field Lines and Flux

  • Magnetic field lines (cont’d)
slide18

B

Area A

B

B

Magnetic Field Lines and Flux

  • Magnetic flux

magnetic flux through a surface

slide19

Magnetic Field Lines and Flux

  • Magnetic flux (cont’d)

Units:

A=C/s, T=N/[C(m/s)]

-> Tm2=Nm/[C/s]=Nm/A

  • Gauss’s law for magnetism

No magnetic monopole has been observed!

slide20

Motion of Charged Particles in a Magnetic Field

  • Case 1: Velocity perpendicular to magnetic field

υ perpendicular to B

The particle moves at constant speed υ in a circle in the plane perpendicular to B.

F/m = a provides the acceleration to the center, so

v

R

F

B

x

slide21

Motion of Charged Particles in a Magnetic Field

  • Case 1: Velocity perpendicular to magnetic field
slide22

Motion of Charged Particles in a Magnetic Field

  • Case 1: Velocity perpendicular to magnetic field

Velocity selector

slide23

Motion of Charged Particles in a Magnetic Field

  • Case 1: Velocity perpendicular to magnetic field

Mass spectrometer

slide24

Motion of Charged Particles in a Magnetic Field

  • Case 1: Velocity perpendicular to magnetic field

Mass spectrometer

slide25

Motion of Charged Particles in a Magnetic Field

  • Case 1: Velocity perpendicular to magnetic field

Mass spectrometer

slide26

Motion of Charged Particles in a Magnetic Field

  • Case 1: Velocity perpendicular to magnetic field

Mass spectrometer

slide27

Motion of Charged Particles in a Magnetic Field

  • Case 2: General case (vat any angle to B)
slide28

Motion of Charged Particles in a Magnetic Field

  • Case 2: General case (cont’d)

Since the magnetic field does not exert

force on a charge that travels in its direction,

the component of velocity in the magnetic

field direction does not change.

slide29

Magnetic Force on a Current-Carrying Conductor

  • Magnetic force on a current (straight wire)
slide30

Magnetic Force on a Current-Carrying Conductor

  • Magnetic force on a current (straight wire) (cont’d)
slide31

Magnetic Force on a Current-Carrying Conductor

  • Magnetic force on a current (curved wire)
slide32

Magnetic Force on a Current-Carrying Conductor

  • Magnetic force on a current: Example1
slide33

Magnetic Force on a Current-Carrying Conductor

  • Magnetic force on a current: Example1 (cont’d)
slide34

Magnetic Force on a Current-Carrying Conductor

  • Magnetic force on a current: Example1 (cont’d)
slide35

Magnetic Force on a Current-Carrying Conductor

  • Magnetic force on a current: Example1 (cont’d)
slide36

Magnetic Force on a Current-Carrying Conductor

  • Magnetic force on a current: Example1 (cont’d)
slide37

Magnetic Force on a Current-Carrying Conductor

  • Magnetic force on a current: Example2
slide38

Magnetic Force on a Current-Carrying Conductor

  • Magnetic force on a current: Example2 (cont’d)
slide39

Magnetic Force on a Current-Carrying Conductor

  • Magnetic force on a current: Example2 (cont’d)
slide40

Force and Torque on a Current Loop

  • Plane of loop is parallel to the magnetic field
slide41

Force and Torque on a Current Loop

  • Plane of loop : general case

if q=90o

slide42

Force and Torque on a Current Loop

  • Plane of loop and magnetic moment
slide43

Force and Torque on a Current Loop

  • Plane of loop : magnetic moment (cont’d)

The same magnetic dipole moment formulae work for any shape of planar loop.

Any such loop can be filled by a rectangular mesh as in the sketch. Each sub-loop is made to carry the current NI. You will now see that all the interior wires have zero current and are of no consequence. Nevertheless, each sub-loop contributes to μ in proportion to its area.

slide44

Force and Torque on a Current Loop

  • Plane of loop : magnetic moment (cont’d)
slide45

Force and Torque on a Current Loop

  • Potential energy of a magnetic dipole

Work done by the torque when the magnetic

moment is rotated by df :

In analogy to the case of an electric dipole

in Chapter 22, we define a potential energy:

Potential energy of a magnetic

dipole at angle f to a magnetic

field

slide46

Current increased

  • μ=I • Area increases
  • Torque fromBincreases
  • Angle of needle increases

Current decreased

  • μdecreases
  • Torque fromBdecreases
  • Angle of needle decreases

Applications

  • Galvanometer

We have seen that a magnet can exert a torque on a loop of current – aligns the loop’s “dipole moment” with the field.

  • In this picture the loop (and hence the needle) wants to rotate clockwise
  • The spring produces a torque in the opposite direction
  • The needle will sit at its equilibrium position
slide47

Applications

  • Motor

Slightly tip the loop

Restoring force from the magnetic

torque

Oscillations

Now turn the current off, just as the loop’s μ is aligned with B

Loop “coasts” around until itsμ is ~antialigned withB

Turn current back on

Magnetic torque gives another kick to the loop

Continuous rotation in steady state

slide48

VS I

t

Applications

  • Motor (cont’d)
  • Even better
  • Have the current change directions every half rotation
  • Torque acts the entire time
  • Two ways to change current in loop:
    • Use a fixed voltage, but change the circuit (e.g., break connection every half cycle
    •  DC motors
    • 2. Keep the current fixed, oscillate the source voltage
  • AC motors
slide49

Applications

  • Motor (cont’d)

flip the current direction

slide50

Applications

charges accumulate

(in this case electrons)

  • Hall effect

-

-

-

+

+

+

Measuring

Hall voltage

(Hall emf)

In a steady state

qEH =qvdB

Charges move sideways until the Hall field EH grows to balance the force due to the magnetic field:

n can be measured

slide51

Applications

  • Electromagnetic rail gun
slide52

v

1

x

x

r

x

x

2

Exercises

  • Exercise 1

If a proton moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what is the speed of the proton and the frequency of motion?

slide53

F

v

N

B

Exercises

  • Exercise 2

Example of the force on a fast moving proton due to the earth’s magnetic field. (Already we know we can neglect gravity, but can we neglect magnetism?)

Let v = 107 m/s moving North.

What is the direction and magnitude of F?

Take B = 0.5x10-4 T and v B to get maximum effect.

(a very fast-moving proton)

vxB is into the paper (west). Check with globe

magnetic field of a moving charge

Magnetic field produced by a moving charge

Magnetic Field of a Moving Charge

Note the factor of μ0 /4π the constant of proportionality needed just as 1/(4πε0) is needed in electrostatics.

slide55

Magnetic Field of a Current Element

  • Magnetic field produced by a current element

ds

For an element ds of a conductor carrying a current I there are n A ds charges with drift velocity υd (using priciple of superposition).

number of charge q

ds

ds

slide56

Magnetic Field of a Current Element

  • Biot-Savart law

ds

Note: ds is dL in the textbook.

ds

Note that ds is in the direction of I, but has a magnitude which is ds the length of wire considered.

Deduced by Biot and Savart c. 1825 from experiments with coils

slide57

dB

The magnitude of the field dB is:

Total magnetic field at P is found by summing over all the current elements ds in the wire.

Magnetic Field of a Current Element

  • Biot-Savart law (cont’d)

P

r

I

ds

slide58

y

P

r

R

x

ds

I

x

So the magnitude of dB is given by:

Magnetic Field of a Straight Current Carrying Conductor

  • A straight wire of length L
  • A thin straight wire of length Lcarries constant current I .
  • Calculate the total B field at P.

 dB

slide59

y

 dB

P

r

R

x

ds

I

x

Magnetic Field of a Straight Current Carrying Conductor

  • A straight wire of length L (cont’d)
slide60

y

 dB

P

r

R

x

ds

I

x

Magnetic Field of a Straight Current Carrying Conductor

  • A straight wire of length L (cont’d)

In the limit (L/R) →∞

Magnetic field by a long straight wire

slide61

Magnetic Field of a Straight Current Carrying Conductor

  • A straight wire of length L (cont’d)

B

B

B

I

B

slide62

Magnetic Field of a Straight Current Carrying Conductor

  • Example: A long straight wire

Iron filings

slide63

The magnetic field due to segments A´A and CC´ is zero because ds is parallel to along these paths.

Along path AC, dsand are perpendicular.

Magnetic Field of a Current Element

  • Example

Calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two straight segments and a circular arc of radius R that subtends angle .

A

ds

I

C

R

O

Note: B field at the centre of a loop, =2

slide64

Force Between Parallel Conductors

  • Two parallel wires

At a distance afrom the wire with current I1the magnetic field due to the wire is given by

slide65

Force Between Parallel Conductors

  • Two parallel wires (cont’d)

Parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other.

slide66

Force Between Parallel Conductors

  • Definition of ampere

The chosen definition is that for a = L = 1m, The ampere is made to be such that F2 = 2×10−7 N when I1=I2=1 ampere 

This choice does two things (1) it makes the ampere (and also the volt) have very convenient magnitudes for every day life and (2) it fixes the size of μ0 = 4π×10−7. Note ε0 = 1/(μ0c2). All the other units follow almost automatically.

slide67

Use to find B field at the center of a loop of wire.

I

R

First find

is a vector coming out of the paper at the same angle anywhere on the circle. The angle is constant.

R

i

Magnetic Field of a Circular Current Loop

  • Magnetic field produced by a loop current

Loop of wire lying in a plane. It has radius R and total current I flowing in it.

Magnitude of B field at center of loop. Direction is out of paper.

slide68

I

Magnetic Field of a Circular Current Loop

  • Example 1:

Loop of wire of radius R = 5 cm and current I = 10 A. What is B at the center? Magnitude and direction

Direction is out of the page.

slide69

I

Total length of arc is S.

0

R

where S is the arc length S =R0

0 is in radians (not degrees)

Magnetic Field of a Circular Current Loop

  • Example 2:

What is the B field at the center of a segment or circular arc of wire?

P

Why is the contribution to the B field at P equal to zero from

the straight section of wire?

slide70

Ampere’s Law

  • Ampere’s law : A circular path
  • Consider any circular path of radius R centered on the wire carrying current I.
  • Evaluate the scalar product B·ds around this path.
  • Note that B and ds are parallel at all points along the path.
  • Also the magnitude of B is constant on this path. So the sum of all the B·ds terms around the circle is

Previously from the Biot-Savart’s law we had

Ampere’s Law

On substitution for B

slide71

z

Ampere’s Law

^

k

  • Ampere’s law : A general path

^

y

r

^

q

q

x

Let us look at the integral along any shape of closed path in 3D. The most general ds is

Where unit vectors are used for the radial r and the tangential directions q and for z along the wire k. In this system we have

^

^

^

tangential component of ds

For any path which encloses the wire

For any path which does not enclose the wire

slide72

Ampere’s Law

  • Ampere’s law :
  • This law holds for an arbitrary closed path that is threaded by a
  • steady current.
  • I is the total current that passes through a surface bounded by the
  • closed path.
slide73

Ampere’s Law

  • Electric field vs. magnetic field
  • Electric Field
  • General: Coulomb’s Law
  • High symmetry: Gauss’s Law
  • Magnetic Field
  • General: Biot-Savart Law
  • High Symmetry: Ampère’s Law
slide74

Applications of Ampere’s Law

  • Magnetic field by a long cylindrical conductor

A long straight wire of radius R carries a steady current I that is uniformly distributed through the cross-section of the wire. Outside R.

In region where r< Rchoose a circle of radius r centered on the wire as a path of integration. Along this path, B is again constantin magnitude and is always parallel to the path.

  • Now Itot≠I.
  • However, current is uniform over the cross-section of the wire.
  • Fraction of the current I enclosed by the circle of radius r < R equals the ratio of the area of the circle of radius r and the cross section of the wire R2.
slide75

B

R

r

Applications of Ampere’s Law

  • Magnetic field by a long cylindrical conductor
slide76

Applications of Ampere’s Law

  • Magnetic field by a circular current

Consider circular current carrying loop. Calculate B field at point P, a dist x from the centre of the loop on the axis of the loop.

Ids

ds

Again in this case vector I dsis tangent to loopand perp to vector r from current element to point P. dB is in direction shown, perp to vectors r and I ds. Magnitude dB is:

ds

ds

slide77

Applications of Ampere’s Law

  • Magnetic field by a circular current (cont’d)

ds

ds

Ids

  • Integrate around loop, all components of dB perp to axis (e.g. dBy). integrate to zero.
  • Only dBx , the components parallel to axis contribute.

ds

Field due to entire loop obtained by integrating:

slide78

Applications of Ampere’s Law

  • Magnetic field by a circular current (cont’d)

But I, R and x are constant

ds

ds

Ids

B on the axis of a current loop

slide79

Applications of Ampere’s Law

  • Magnetic field by a circular current (cont’d)

x >>R

Limits: x 0

Compare case of electric field on axis of electric dipole far from dipole

vs.

slide80

Applications of Ampere’s Law

  • Magnetic field by a solenoid

When the coils of the solenoid are closely spaced, each turn can be regarded as a circular loop, and the net magnetic field is the vector sum of the magnetic field for each loop. This produces a magnetic field that is approximately constant inside the solenoid, and nearly zero outside the solenoid.

I

slide81

Applications of Ampere’s Law

  • Magnetic field by a solenoid (cont’d)

The ideal solenoid is approached when the coils are very close together and the length of the solenoid is much greater than its radius. Then we can approximate the magnetic field as constant inside and zero outside the solenoid.

I

slide82

Applications of Ampere’s Law

  • Magnetic field by a solenoid (cont’d)

Use Ampère’s Law to find B inside an ideal solenoid.

slide83

N is the number of loops in the toroid, and I is the current in each loop

Applications of Ampere’s Law

  • Magnetic field by a toroid

A toroid can be considered as a solenoid “bent” into a circle as shown. We can apply Ampère’s law along the circular path inside the toroid.

slide84

Exercises

  • Problem 1

The wire semicircles shown in Fig. have radii a and b. Calculate the net

magnetic field that the current in the wires produces at point P.

I

Since point P is located at a symmetric position

with respect to the two straight sections where the

current I moves (anti)parallel to the radial direction.

So there is no contributions from these segments.

The contribution from the semicircle of radius a is

a half of that from a complete circle of the same radius:

Similarly the contribution from the semicircle of radius b is:

From principle of superposition, the net magnetic field at point P is:

I

P

slide85

Exercises

  • Problem 2

Long, straight conductors with square cross sections and

each carrying current I are laid side-by-side to form an

infinite current sheet. The conductors lie in the xy-plane,

are parallel to the y-axis and carry current in the +y

direction. There are n conductors per unit length measured

along the x-axis. (a) What are the magnitude and direction

of the magnetic a distance a below the current sheets?

(b) What are the magnitude and direction of the magnetic

field a distance a above the current sheet?

x

z

y

slide86

Exercises

  • Problem 2 (cont’d)

B

  • Below sheet, all the magnetic
  • field contributions from different
  • wires add up to produce a magnetic
  • field that points in the positive x-direction. Components in the z-direction
  • cancel. Using Ampere’s law, where we use the fact that the field is anti-
  • symmetric above and below the current sheets, and that the legs of the path
  • perpendicular provide nothing to the integral. So, at a distance a beneath the
  • sheet the magnetic field is:
  • b) The field has the same magnitude above the sheet, but points in the negative
  • x-direction.

B

L