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PHY 102. (F

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Good day. Here we go …………………..Good day. Here we go …………………..Good day. Here we go …………………..Good day. Here we go …………………..

PHY102- GENERAL PHYSICS II

Text Book: Fundamentals of Physics

Authors: Halliday, Resnick & Walker

Edition: 8th Extended

Lecture Schedule

10/24/2012Lecture Schedule

TOPICS: Dates

Ch. 28 – Magnetic Fields 12th & 15th October

Ch. 29 – Magnetic Fields due

to Currents 17th & 19th October

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PHY 102. (F

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Date for 1st Test: 10 November, 2012 (Tentative)

MR. O.G. FAWOLEMR. O.G. FAWOLEMR. O.G. FAWOLEMR. O.G. FAWOLERm. G14 F {Dept. of Physics.}

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CHAPTER 28. MAGNETIC FIELD

28.1. What is Physics?28.2. What Produces a Magnetic Field?28.3. The Definition of Magnetic Field P

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28.3. The Definition of Magnetic Field 28.4. Crossed Fields: Discovery of the Electron28.5. Crossed Fields: The Hall Effect28.6. A Circulating Charged Particle28.7. Cyclotrons and Synchrotrons28.8. Magnetic Force on a Current-Carrying wire28.9. Torque on a Current Loop

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28.9. Torque on a Current Loop28.10. The Magnetic Dipole Moment

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1. MAGNETIC FIELDS1. MAGNETIC FIELDS1. MAGNETIC FIELDS1. MAGNETIC FIELDS

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SUBSUBSUBSUB----SECTIONSSECTIONSSECTIONSSECTIONS

� Magnetic Fields.

You must understand the similarities and differences between electric fields and field lines, and magnetic fields and field lines.

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� Magnetic Force on Moving Charged Particles.

You must be able to calculate the magnetic force on moving charged particle.

� Magnetic Flux and Gauss’ Law for Magnetism.

You must be able to calculate magnetic flux and recognize the

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You must be able to calculate magnetic flux and recognize the consequences of Gauss’ Law for Magnetism.

� Motion of a Charged Particle in a Uniform Magnetic Field.

You must be able to calculate the trajectory and energy of a charged particle moving in a uniform magnetic field.

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Recall how there are two kinds of electrical charge (+ and -), and likes repel opposites attract.

MAGNETISMMAGNETISMMAGNETISMMAGNETISM

Similarly, there are two kinds of magnetic poles (north and south), and like poles repel, opposites attract.

10/24/2012poles repel, opposites attract.

S N S NAttract

S NRepel

SN

N

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Repel Attract

S N

S N

S N

SN6

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THE MAGNETIC FIELDPHY 102. (F

AWOLE O.G.)� A magnetic field exists in the region around a

magnet. The magnetic field is a vector that has both

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magnet. The magnetic field is a vector that has both magnitude and direction.

� The direction of the magnetic field at any point in space is the direction indicated by the north pole

of a small compass needle placed at that point.

MAGNETIC FIELD.MAGNETIC FIELD.MAGNETIC FIELD.MAGNETIC FIELD.

The earth has associated with it a magnetic field, with poles near the geo-graphic poles. The pole of a magnet attracted to the earth’s north geographic pole is the magnet’s north pole.

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Npole.

The pole of a magnet attracted to the earth’s south geographic pole is the magnet’s south pole.

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N

S

There are two ways to produce magnetic field namely:� Using moving charged particles, such as a current in a wire, to make an electromagnet� Using elementary particles such as electron, because they

Just as we used the electric field to help us “explain” and visualize electric forces in space, we use the magnetic field to help us “explain” and visualize magnetic forces in space.

Magnetic field lines point in the same direction that the north pole of a compass would point.

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� Using elementary particles such as electron, because they have intrinsic magnetic fields around them

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THE MAGNETIC FIELD LINEPHY 102. (F

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� The lines originate from the north pole and end on the south pole; they do not start or stop in mid-space.

� The magnetic field at any point is tangential to the magnetic field line at that point.

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� The magnetic field at any point is tangential to the magnetic field line at that point.

� The strength of the field is proportional to the number of lines per unit area that passes through a surface oriented perpendicular to the lines.

� The magnetic field lines will never come to cross each other.

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WHAT PRODUCES A MAGNETIC FIELD?PHY 102. (F

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� Moving electrically charged particles, such as a current, produce a magnetic field

� Permanent magnet. Elementary particles such as electrons have an intrinsic magnetic field around them. The magnetic

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have an intrinsic magnetic field around them. The magnetic fields of the electrons in certain materials add together to give a net magnetic field around the material. Such addition is the reason why a permanent magnet has a permanent magnetic field. In other materials, the magnetic fields of the electrons cancel out, giving no net magnetic field surrounding the material

THE DEFINITION OF MAGNETIC FIELD, B10/24/2012

We use the symbol B for magnetic field.

Remember: magnetic field lines point away from north poles, and towards south poles.

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S NS N

The SI unit for magnetic field is the Tesla.

Remember, units of field are force per “something.”

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The earth’s magnetic field has a magnitude of roughly 0.5 G, or 0.00005 T. A powerful permanent magnet, like the kind you might find in headphones, might produce a magnetic field of 1000 G, or 0.1 T. Superconducting magnets can produce a field of over 10 T.

Where G (gauss) is an old unit of magnetic field. 1 tesla(T) = 104 gauss(G).

MAGNETIC FIELD, B.

To determine the electric field E at a point, a test particle of charge, q at rest is placed at the point. The electric force, FE acting on the particle can then be measured. E is then given by: 1

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(1)

We cannot define magnetic field, like this because a magnetic monopole is not available. Hence, we have to define , in another way, in terms of the magnetic force, FB exerted on a moving electrically charged test particle.

Magnetic field, B can be said to be a vector quantity that is directed along the zero-force axis. When the particle velocity, v, is directed perpendicular to the direction of the zero-force axis, the magnetic field is given as:

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(2)

where q is the charge of the particle.

12A charged particle moving in a magnetic field experiences a force. The magnetic force equation predicts the effect of a magnetic field on a moving charged particle.

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MAGNETIC FORCE ON A CHARGED PARTICLE

When a charge is placed in a magnetic field, it experiences a magnetic force if two conditions are met:

1. The charge must be moving. No magnetic force acts PHY 1

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1. The charge must be moving. No magnetic force acts on a stationary charge.

2. The velocity of the moving charge must have a component that is perpendicular to the direction of the field.

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� ��F=qv B

force on particle

velocity of charged particle

magnetic field vector

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particle

What is the magnetic force if the charged particle is at rest?

What is the magnetic force if v is (anti-)parallel to B? The force FB on the particle is equal to the charge q times the cross product of its velocity v, and the magnetic field B (all measured in the same frame of reference).

RECALL: A X B = ABsinθ. Hence;

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RECALL: A X B = ABsinθ. Hence; (3)

where θ is the angle between the directions of velocity, v and magnetic field, B

θ ⊥ ⊥F= q vB sin = q v B= q vB

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MAGNETIC FORCE ON A CHARGED PARTICLEPHY 102. (F

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Right-Hand Rule

The force acting on a charged particle moving with velocity through a magnetic field is always

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Fingers out in direction of velocity, thumb perpendicular to them. rotate your hand until your palm points in the direction of

magnetic field. Thumb points in direction of magnetic force on + charge.

particle moving with velocity through a magnetic field is alwaysperpendicular to and .

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“Foolproof” technique for calculating both magnitud e and direction of magnetic force.

� ��F=qv B

ˆˆ ˆ �

i j k

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hints for solving problemshints for solving problemshints for solving problemshints for solving problems:

�

x y z

x y z

F = q d e t v v v

B B B

θF= q vB sin�A charged particle moving along or opposite to the direction of a

magnetic field will experience no magnetic force.

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magnetic field will experience no magnetic force.�Conversely, the fact that there is no magnetic force along some direction

does not mean there is no magnetic field along or opposite to that direction.

� It’s OK to use if you know that v and B are perpendicular, or you are calculating a minimum field to produce a given force (understand why).

F= q vB

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MODEL PROBLEMS.MODEL PROBLEMS.MODEL PROBLEMS.MODEL PROBLEMS.

1. An electron that has a velocity, = (2.0 x 106 m/s) i + (3.0 x 106 m/s) j moves through the

uniform magnetic field, = (0.030 T)i – (0.15 T)j.(a). Find the force on the electron due to the magnetic field.(b). Repeat your calculation for a proton having the same velocity.

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SOLUTION.

1(a). The force on the electron is

= (-1.6 x 10-19 C){(2.0 x 106 m/s)(-0.15 T) – (3.0 x 106 m/s)(0.030 T)}

= (6.2 x 10-14 N)The magnitude of is 6.2 x 1014 N and it points in the direction of the positive

z direction(b) This implies repeating the calculations in (a) with a change in the sign of the

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z direction(b) This implies repeating the calculations in (a) with a change in the sign of the charge.

Hence, has the same magnitude but points in the negative z direction

= - (6.2 x 10-14 N)

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2. A proton travelling at 23.0º with respect to the direction of a magnetic field of strength 2.60 mT experiences a magnetic force of 6.50 x 10-17 N. Calculate (a) the proton speed and (b) its kinetic energy in eV.

SOLUTION.

1 (a). From eqn. (3)

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(b). The kinetic energy of the proton is

K.E=

Since 1.6 x 10-19 J = 1 eV1.34 x 10-16 J = 835 eV

PRACTICE

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PRACTICE

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2 (a). A proton is moving with a velocity v = v 0j in a region of uniform magnetic field. The resulting force is F = F 0i. What is the magnetic field (magnitude and direction)?

(b) In the above example, what is the minimum magne tic field that can be present (magnitude and direction)?

PHY 102. (F

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USEFUL HINTS

θF= q vB sin

� A charged particle moving along or opposite to the direction of a magnetic field will experience no magnetic force.

� Conversely, the fact that there is no magnetic forc e along some direction does not mean there is no magnetic field along or o pposite to that direction.

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does not mean there is no magnetic field along or o pposite to that direction.

� It’s OK to use if you know that v an d B are perpendicular, or you are calculating a minimum field to produce a given forc e (understand why).

F = q vB

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DIFFERENCES OF ELECTRIC AND MAGNETIC FIELDS

1. Direction of forces� The electric force on a charged particle

(both moving and stationary) is always parallel (or anti-parallel) to the electric field P

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(both moving and stationary) is always parallel (or anti-parallel) to the electric field direction.

� The magnetic force on a moving charged particle is always perpendicular to both magnetic field and velocity of the particle. No magnetic force on a stationary charged particle.

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particle.

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2. THE WORK DONE ON A CHARGED PARTICLE:� The electric force can do work on the particle.

� The magnetic force cannot do work and change the kinetic energy of the charged particle.

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� Both an electric field, E and magnetic field, B can produce a force on a charged particle. When the two fields are perpendicular to each

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CROSSED FIELDS: DISCOVERY OF THE ELECTRON

When the two fields are perpendicular to each other, they are said to be cross fields.

� Studying the crossed field using J.J. Thomson’s expt. that led to the discovery of electrons.

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Cathode ray tube

CROSSED FIELDS CONTD.

In the arrangement the figure above, electron are forced up the page by the electric field and down the page by magnetic field meaning the forces are in opposition.

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in opposition.

Thomson’s procedure was equivalent to the following series of steps.

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1. Set and to zero and note the position of the spot on screen S due to the undeflected beam.2. Turn on and measure the resulting beam deflection y.

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deflection y.3. Maintaining , now turn on and adjust its value until the beam returns to the undeflected position. (With the forces in opposition, they can be made to cancel.)

CROSSED FIELDS CONTD.

The deflection of the particle at the far end of

the plate is:

(4)

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Where v is the particle’s speed, m its mass and q its charge, L is the length of the plates.

When the two fields are adjusted so that the two deflecting forces cancel (step in 3) , then eqn (1) = eqn (3)

qE = qvB sin90 = qvB

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qE = qvB sin90 = qvB

v =

Substituting for v in eq. (4):

(5)

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CROSSED FIELDS: THE HALL EFFECT

Can the drifting conduction electrons in a copper wire also be deflected by a magnetic field? In 1879, E.H Edwin showed they

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In 1879, E.H Edwin showed they can.

The Hall effect allows us to find out whether the charge carriers in a conductor are positively or negatively charged. The number of such carriers per unit volume

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of such carriers per unit volume of the conductor can also be determined.

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CROSSED FIELDS: THE HALL EFFECT

Can the drifting conduction electrons in a copper wire also be deflected by a magnetic field?

A Hall potential difference V of magnitude V = Ed is associated with the electric field across strip width d P

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A ld=

associated with the electric field across strip width dWhen the electric and magnetic forces are in balance

dE v B=

The drift speed,

Where J is the charge density in the strip, A is the cross-sectional area of the strip and n is the number density of

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A ld=

V iB

d neld=

of the strip and n is the number density of charge carriers i.e. number per unit volume, � =A/d is the thickness of the strip.

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EXAMPLE.A metal strip 6.50 cm long, 0.850 cm wide and 0.760 mm thick moves with constant velocity, v through a uniform magnetic field B = 1.20 mT directed perpendicular to the strip as shown in

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perpendicular to the strip as shown in the figure. A potential difference of 3.90 nV is measured between point x and y across the strip. Calculate the speed v.

Solution

For a free charge q inside the metal strip with velocity v, we have

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we have

We set this force equal to zero and use the relation between (uniform) electric field and potential difference. Thus,

= 3.82 x 10-4 m/s.

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THE MOTION OF A CHARGED PARTICLE IN A CONSTANT

MAGNETIC FIELD

If a particle moves in a circular path at a constant speed, it is sure that the net force acting on the particle is constant in P

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the particle is constant in magnitude and direction, pointing towards the centre of the circle, always perpendicular to the particle’s velocity.

Applying Newton’s 2nd law to uniform circular motion, we have:

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uniform circular motion, we have:

(6)

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The force acting on the particle has magnitude of |q|VB, hence

(7)

(8)

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Note that T, f and ω do not depend on the speed of the particle (provided the speed is less than the speed of light).

HELICAL PATHS

Frequency f =

The angular frequency,

(9)

(10)

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HELICAL PATHS

When the velocity of the charged particle has a component parallel to the magnetic field, the particle will move in a helical path about the direction of the field vector.

� Read more about motion of charged particle in helical paths.

PRACTICE QUESTION10/24/2012

1. A proton is released from rest at point A, which is located next to the positive plate of a parallel plate capacitor (see Figure 28a). The proton then accelerates toward the negative plate, P

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accelerates toward the negative plate, leaving the capacitor at point B through a small hole in the plate. The electric potential of the positive plate is 2100 V greater than that of the negative plate, so VA–VB = 2100 V. Once outside the capacitor, the proton travels at a constant velocity until it enters a region of constant magnetic field of magnitude

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of constant magnetic field of magnitude 0.10 T. The velocity is perpendicular to the magnetic field, which is directed out of the page in Figure 28a. Find (a) the speed vB of the proton when it leaves the negative plate of the capacitor, and (b) the radius r of the circular path on which the proton moves in the magnetic field.

Fig. 28a

CYCLOTRON AND SYNCHROTRON

The number of revolutions per unit time is f = ω/2π. This frequency f is independent of the radius R of the path. It is called the cyclotron frequency; in a particle

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cyclotron frequency; in a particle accelerator called a cyclotron, particles moving in nearly circular paths are given a boost twice each revolution, increasing their energy and their orbital radii but not their angular speed or

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not their angular speed or frequency.

(11)

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For resonance condition:

(12)

MAGNETIC FORCE ON A CURRENT-CARRYING WIRE

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Fig. 28c explains what happens in Fig. 28b. A conduction electron drifting downwards with an assumed drift speed . A force of magnitude must act on such electron. This force must be directed to the RIGHT. P

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Fig. 28b

RIGHT.

Consider a length L of the wire in fig. 28c. All the conduction electrons in this section of the wire will drift past the xx plane in a time t =

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Fig. 28c

(13)eq. (13) gives the magnetic force acting on a length L of straight wire carrying a current i and immersed in a magnetic field B that is perpendicular to the wire.

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If the magnetic field is notperpendicular to the wire, as in

MAGNETIC FORCE ON A CURRENT-CARRYING WIRE CONTD.PHY 102. (F

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perpendicular to the wire, as in Fig. 28d, the magnetic force is given by

(14)Define as a length vector that has magnitude L and is directed along the wire segment in the direction of the (conventional) current.

Fig. 28d

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FB is always perpendicular to the plane defined by vectors L and B as shown in fig. 28d.If a wire is not straight or the field is not uniform. We can imagine the wire broken up into small straight segments and then apply:

(15)

In the differential limit we can write: (16)

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SAMPLE PROBLEM

A straight, horizontal length of copper wire has a current i=28 A flowing through it. What are the magnitude and direction of the minimum magnetic field needed to suspend the wire—that is, to P

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field needed to suspend the wire—that is, to balance the gravitational force on it? The linear density (mass per unit length) of the wire is 46.6 g/m.

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0netF =�

TORQUE ON A CURRENT LOOP

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The force F4 has the same magnitude as force F2 but they act in opposite directions, hence they cancel out exactly. The vector , which is normal to the plane of the coil with direction determined by right hand rule shown in fig. (b).

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siniABτ θ′ =A(=ab) is area of loop

(15)

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When a current-carrying loop is placed in a magnetic field, the loop tends to rotate such that its normal becomes aligned with the magnetic field

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TORQUE ON A CURRENT COIL

A coil containing N loops, each of area A, the force on each side is N times larger, and the torque on

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on each side is N times larger, and the torque on the coil becomes:

(16)

Where NiA are properties of the coil: its number of turns, its area and the current it carries

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turns, its area and the current it carries

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THE MAGNETIC DIPOLE MOMENT, OF A COIL.

To account for the torque on the coil due to the magnetic field, a magnetic dipole moment is assigned. The magnitude of is given by:

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(17)

Recall: For electric field, torque exerted by an electric field on an electric dipole is given by:

The magnetic potential energy

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The magnetic potential energy(18)

•If an applied torque rotates a magnetic dipole from an initial orientation to another orientation , then the w ork Wa done on the dipole by the applied torque is

(19)

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Model QuestionA circular wire loop of radius 15.0 m carries a current of 2.0 A. It is placed so that the normal to its plane makes an angle of 41.0o with a uniform magnetic field of magnitude 12.0 T. (a) Calculate the magnitude of the magnetic dipole moment of

the loop. (b) What is the magnitude of the torque acting on the loop? P

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the loop. (b) What is the magnitude of the torque acting on the loop?

Solution

(a)

= 0.184 A.m2

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(b) Torque τ =|µ x B| = µBsinθ = (0.184Am2)(12.0 T) sin 41.0o

= 1.45 Nm

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CHAPTER 29. MAGNETIC FIELD DUE TO CURRENTS

29.1. What is Physics?29.2. Calculating the Magnetic Field Due to a P

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29.2. Calculating the Magnetic Field Due to a Current

29.3. Force Between Two Parallel Currents29.4. Ampere's Law29.5. Solenoids and Toroids29.6. A Current-Carrying Coil as a Magnetic Dipole

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29.6. A Current-Carrying Coil as a Magnetic Dipole

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WHAT IS PHYSICS?

A moving charged particle produces a magnetic field around itself P

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around itself

Activation of regions in the human brain by certain activities such as reading produces

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reading produces magnetic field that can be detected by several devices.

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MAGNETIC FIELD DUE TO A CURRENT

� Quantitative rule for computing

Jean-BaptisteBiot (1774-1862)

Felix Savart (1791-1841)

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� Quantitative rule for computing the magnetic field from any electric current

� Choose a differential element of wire of length ds and carrying a current i

� The field dB from this element at

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� The field dB from this element at a point located by the vector r is given by the Biot-Savart Law in scalar form:

µµµµ0 =4ππππx10–7 T•m/A=1.26 x 10-6 T.m/A(permeability constant)

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BIOT-SAVART LAW

� An infinitely long straight wire carries a

current i.

� Determine the magnetic field generated at

a point located at a perpendicular distance PH

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a point located at a perpendicular distance

R from the wire.

� Choose an element ds as shown

� Biot-Savart Law: dB points INTO the

page. This is the vector form of the Biot-

Savart law.

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� Integrate over all such elements

This law can be used to calculate thenet

magnetic field B produced at a point by

various

Where θ is the angle between the directions of ds and r, a unit vector that points from ds towards P.

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MAGNETIC FIELD DUE TO A CURRENT IN A LONG STRAIGHT WIRE

02

sin

4

i dsdB

r

µ θπ

= i

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RIGHT-HAND RULE

Grasp the element in your right hand with your P

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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

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field lines due to that element.

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MAGNETIC FIELD DUE TO A CURRENT IN A CIRCULAR ARC OF WIRE

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2φ π=

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A CURRENT-CARRYING COIL AS A MAGNETIC DIPOLE

0

2c

iB

R

µ=

For a loop, ϕ=2π, at the center of the loopPH

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470

2 2 3/2( )

2 ( )B z

R z

µ µπ

=+ If

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SAMPLE PROBLEM

The wire in Fig. 29-8a carries a current i and consists of a circular arc of radius R and central angle rad, and two straight sections whose extensions intersect the center C of the arc. What magnetic field does the current produce at C ? P

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current produce at C ?

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TWO CURRENT-CARRYING WIRES EXERT MAGNETIC

FORCES ON ONE ANOTHERPH

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� Two long parallel wire carrying currents exert forces on each other.

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Two long parallel wire carrying currents exert forces on each other.

� To find the force on a current-carrying wire due to a second current-carrying wire, first find the field due to the second wire at the site of the first wire. Then find the force on the first wire due to that field.

� Parallel currents attract each other, and anti-parallel currents repel each other.

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To find the force on wire b due to the current in wire a. That current produces a magnetic field Ba and it is this magnetic field that produces the force on wire b.

(1)

The force, produced on wire b due to Ba: PH

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The force, produced on wire b due to Ba:

(2)

The vectors L and B are perpendicular to each other, hence:

(3)The direction of is the direction of the cross product

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The direction of is the direction of the cross product

The Ampere, A, is that constant current which, if maintained in two straight, parallel conductors of infinite length, of negligible circular cross section, and placed 1 m apart in a vacuum, would produce on each of these conductors a force of magnitude of wire length.

AMPERE'S LAW

� The loop on the integral sign means that is to be integrated around a closed loop, called an Amperian loop. The current ienc is the net current encircled by that closed loop.

� Curl your right hand around the Amperian loop, with the fingers pointing in the direction of integration. A current through the loop in 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.

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Ampère’s Law

0B ds Iµ⋅ =∫� �

�

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Using arguments similar to the ones use for Gauss’s Law, we can show that this result

•Is independent of the shape of the curve around the current.

•Is independent of where the current passes through the curve.

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curve.

•Depends only on the total amount of current through the area enclosed by the integration path.

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012

Magnetic Field inside a current-carrying wire

2through circleI JA r Jπ= =

PH

Y 1

02. (F

AW

OLE O

.G.)

2rµ ∫� �

Assume a uniform current density,

with I the total current passing

through wire of radius R.

2

IJ

Rπ=

22

through 2

rI r J I

Rπ⇒ = =

53

20 2

0 through 2 02

2

2

rB ds I I r

R rB IR

B ds Bl rB

µµ µππ

⋅ = =

⇒ =⋅ = =

∫

∫

� �

� �

�

�

022

IB r

R

µπ

= 0Recall: outside.2

IB

r

µπ

=

10/2

4/2

012MAGNETIC FIELD OF A SOLENOID

PH

Y 1

02. (F

AW

OLE O

.G.)

For a long ideal solenoid

54

where n is the number of turns per unit length of the solenoid

h is the (arbitrary) length of the segment from a to b in the figure.

10/2

4/2

012

MAGNETIC FIELD OF A TOROIDPH

Y 1

02. (F

AW

OLE O

.G.)

55

10/2

4/2

012

EXAMPLE

1. A solenoid is 0.50 m long, has three layers of windings of 750 turns each, and carries a current of 4.0 A. What is the magnetic field at the center of the solenoid? P

HY 1

02. (F

AW

OLE O

.G.)

solenoid?

Solution:

56