giambattista college physics chapter 19

Giambattista College Physics Chapter 19

©2020 McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education.

Chapter 19: Magnetic Forces and Fields

19.1 Magnetic Fields.

19.2 Magnetic Force on a Point Charge.

19.3 Charged Particle Moving Perpendicularly to a Uniform Magnetic Field.

19.4 Motion of a Charged Particle in a Uniform Magnetic Field: General.

19.5 A Charged Particle in Crossed E and B Fields.

19.6 Magnetic Force on a Current-Carrying Wire.

19.7 Torque on a Current Loop.

19.8 Magnetic Field due to an Electric Current.

19.9 Ampère’s Law.

19.10 Magnetic Materials.

19.1 Magnetic Fields

A bar magnet is one instance of a magnetic dipole. By dipole we mean two opposite poles.

Look carefully and note that themagnetic field lines do NOT begin and end like electricfield lines, but form closed loops.

They SEEM to begin at the N pole andend at the S pole, but they really reconnect inside. This was not obviousin the beginning of the study of magnets (1400s!) but will be suggested by the magnetic field of a solenoid (Ch 19.8).

EDITORIAL COMMENT:

I’m generally very unhappy with the way that magnetic field lines are drawn, since studentswill USUALLY have to UNLEARN what these diagramsteach. These aren’t that bad (we’ll seeworse examples).

But we will look at this in lab (if you havelab with me) and I’ll try to stress the difference between reality and the imagination of people who draw these diagrams.

Permanent Magnets 1

In an electric dipole, the electric poles are positive and negative electric charges. A magnetic dipole consists of two opposite magnetic poles.

The end of the bar magnet where the field lines emerge is called the north pole, and the end where the lines go back in is called the south pole. Like poles repel one another and opposite poles attract one another.

Ultimately, magnetic fields and forces are created and felt by currents – moving charges.

The magnetic field is represented by the symbol

Permanent Magnets 2

A compass is simply a small bar magnet that is free to rotate.

Any magnetic dipole, including a compass needle, feels a torque that tends to line it up with an external magnetic field.

A magnet need not have only two poles; it must have at least one north pole and at least one south pole.

©GIPhotoStock/Science Source

Permanent Magnets 2

A compass is simply a small bar magnet that is free to rotate, but NOT move.

Any magnetic dipole, including a compass needle, feels a TORQUE that tends to line it up with an external magnetic field.

There is also a net FORCE toward the area of higher magnetic field.

Magnetic Field Lines

If you divide a magnet in half, you get TWO magnets, each with a N and S pole.

Ultimately, if you keep going, you will have a single atom with a N and a S pole.

Ultimately, we will see that these fields are created by loops of current, with a N pole on one side of the current, and a S pole on the other.

NOT by magnetic particles.

Figure used w permission, from:http://www.gonefcon.com/trucktcom/magnets.htm

Does a good job showing what happens when the magnet is divided – the rest, not so good.

No Magnetic Monopoles

Coulomb’s law for electric forces gives the force acting between two point charges—two electric monopoles.

However, as far as we know, there are no magnetic monopoles—that is, there is no such thing as an isolated north pole or an isolated south pole.

There’s no REASON there shouldn’t exist, and some scientists have speculated that they SHOULD. So they are looking for them and will no doubt get a Nobel if they find one.

But science doesn’t care if they SHOULD or SHOULDN’T. Whether they exist is an EMPIRICAL question – and so far, no one’s ever found one.

Magnetic field lines do not begin on north poles and end on south poles: magnetic field lines are always closed loops.

If there are no magnetic monopoles, there is no place for the field lines to begin or end, so they must be closed loops.

(Later, we’ll see that they loop AROUND the electric currents that CREATE the field).

Interpretation of Magnetic Field Lines

Same as the interpretation of electric field lines.

The direction of the magnetic field vector at any point is tangent to the field line passing through that point and is in the direction indicated by arrows on the field line.

The magnetic field is strong where field lines are close together and weak where they are far apart.

More specifically, if you imagine a small surface perpendicular to the field lines, the magnitude of the magnetic field is proportional to the number of lines that cross the surface, divided by the area.

(If that sounds like “flux”, it’s not an accident.)

The Earth’s Magnetic Field

The south pole of the fictitious magnet faces roughly toward geographic north and the north pole of the magnet faces roughly toward geographic south. (Field lines could be better)

19.2 Magnetic Force on a Point Charge 1

In Chapter 16 we defined the electric field as the electric force per unit charge. The electric force is either in the same direction as the electric field or in the opposite direction, depending on the sign of the point charge.

The magnetic force on a point charge is more complicated—it is not the charge times the magnetic field. The magnetic force depends on the point charge’s velocity as well as on the magnetic field.

If the point charge is at rest, there is no magnetic force.

The magnitude and direction of the magnetic force depend on the direction and speed of the charge’s motion.

We have learned about other velocity-dependent forces, such as the drag force on an object moving through a fluid. Like drag forces, the magnetic force increases in magnitude with increasing velocity.

However, the direction of the drag force is always opposite to the object’s velocity, while the direction of the magnetic force on a charged particle is perpendicular to the velocity of the particle.

MAGNITUDE: We defined an electric field to model the force that two charges exert upon one another. If those charges are moving, there is ANOTHER force that behaves differently:The magnitude of the force is proportional to the speed, the charge (incl sign) and the strength of the field. So the force is proportional to the product:

BUT, if the velocity is parallel to the magnetic field, the force is zero AND is a maximum if the velocity is perpendicular to the magnetic field. A function that is zero when and maximum when is the sine function. So, we can summarize that the magnitude of the magnetic force on a moving charge is:

F ~qvB

θ =0θ =90°=π /2

F=qvB sin (θ )

UNITS: W give the unit for the magnetic field it’s own name (unlike the electric field): The units of magnetic field are Tesla (named for the scientist, not the car). Starting with our expression for the force, we can put all the units in asn see how to write a Tesla in terms of other units:

So a Tesla can be written as:

We’ll use this often to keep track of units when we calculate the magnetic field of various current arrangements.

F=qvB sin (θ )[N ]=[C ][m /s ][T ][ ]

[T ]= [N⋅s ][C⋅m]

= [N ][A⋅m]

DIRECTION: As mentioned, the magnetic force is perpendicular to the velocity. There’s no convenient way to define B so that the force is along B (the way we did with the electric field and the way we’d LOVE to do now if we could). As odd as it is, the most convenient way to define B is to make the force ALSO perpendicular to the magnetic field. So you’re essentially defining the force to be perpendicular to the PLANE formed by these two vectors. There are two choices, so HISTORICALLY it was decided to use what is called the “right hand rule”, meaning you use your right hand. I’ll show you several ways to do the “right hand rule” and leave it to you to decide which you find best for you. (You may recall the RHR from the section on angular momentum: CH 8.9)

DIRECTION:

1) Thumb/fingers/palm: thumb along v; fingers along B; – F in direction of palm

2) Fingers curl/thumb: let fingers curl FROM v to B; thumb in direction of F

3) Thumb/index/middle: thumb along v; index along B – middle (perpendicular) is in direction of F

4) Screw/bolt/jar lid: “rotate” v toward B – F is the direction a screw or bolt of jar lid would go if you rotated it in that direction.

Vector Symbols for Perpendicular to Page

Since magnetism is inherently three-dimensional, we often need to draw vectors that are perpendicular to the page.

Magnetic force points INTO plane of slide

Magnetic FIELD points OUT of plane of slide

Vector symbols: • or out of page; or into the page

EXAMPLES:

Use the RHR to find the direction of the magnetic force (which is in the direction of v x B)

EXAMPLES:

Use the RHR to find the direction of the velocity or magnetic field (whichever is missing)

Magnetic force on a moving point charge(summarized as cross product)

Magnitude : BF q B sin

Direction : perpendicular to both and use the right and hand rule;

rule to find , then reverse it if is negative.q

Problem-Solving Technique: Finding the Magnetic Force on a Point Charge 1

1. The magnetic force is zero if (a) the particle is not moving (v = 0), (b) its velocity has no component perpendicular to the magnetic field (v⊥ = 0), or (c) the magnetic field is zero.

2. Otherwise, determine the angle between the velocity and magnetic field vectors when the two are drawn starting at the same point.

3. Find the magnitude of the force from FB = |q|vB sin θ, using the magnitude of the charge (since magnitudes of vectors are nonnegative).

4. Determine the direction of using the right-hand rule. The magnetic force is in the direction of if the charge is positive. If the charge is negative, the force is in the direction opposite to

Work Done by the Magnetic Field

Because the magnetic force on a point charge is always perpendicular to the velocity, the magnetic force does no work.

If no other forces act on the point charge, then its kinetic energy does not change.

The magnetic force, acting alone, changes the direction of the velocity but not the speed (the magnitude of the velocity).

Example 19.1 (1)

Cosmic rays are charged particles moving toward Earth at high speeds.

The origin of the particles is not fully understood, but explosions of supernovae may produce a significant fraction of them.

About seven eighths of the particles are protons that move toward Earth with an average speed of about two thirds the speed of light.

Example 19.1(2)

Suppose that a proton is moving straight down, directly toward the equator.

(a) What is the direction of the magnetic force on the proton due to Earth’s magnetic field?

(b) Explain how Earth’s magnetic field shields us from bombardment by cosmic rays.

(c) Where on Earth’s surface is this shielding least effective?

Example 19.1 Solution 1

BSince and is positive,the

magnetic force is into the page or east.

(b) Without Earth’s magnetic field, the proton would move straight down toward Earth’s surface.

The magnetic field deflects the particle sideways and keeps it from reaching the surface. (They are trapped in what are called the Van Allen belts).

Many fewer cosmic ray particles reach the surface than would do so if there were no magnetic field.

Occasionally (every few hundred THOUSAND years) the magnetic field switches, going through zero. We have no protection whils the magnetic field is zero.

(c) Near the poles, the component of v perpendicular to the field (v⊥) is a small fraction of v.

Since the magnetic force is proportional to v⊥, the deflecting force is much less effective near the poles.

Example 19.2

At a certain place, Earth’s magnetic field has magnitude 0.050 mT. The field direction is 70.0° below the horizontal; its horizontal component points due north.

(a) Find the magnetic force on an oxygen ion (O2−) moving due

east at 250 m/s.

(b) Compare the magnitude of the magnetic force with the ion’s weight, 5.2 × 10−25 N, and to the electric force on it due to Earth’s fair-weather electric field (150 N/C downward).

(a) The ion is moving east; the field has northward and downward components, but no east/west component.

Therefore, v and B are perpendicular; θ = 90° and sin θ = 1.

The magnitude of the magnetic force is then

1 6 10 C 250 m/s 5 0 10 T

2 0 10 N

(a) continued.

Since the ion is negatively charged, the magnetic force is in the direction opposite to ; it is 20.0° below the horizontal, with its horizontal component pointing south.

The magnetic force on the ion is much stronger than the gravitational force and much weaker than the electric force.

19 17E 1 6 10 C 150 N/C 2 4 10 NF q E

Example 19.3

An electron moves with speed 2.0 × 106 m/s in a uniform magnetic field of 1.4 T directed due north. At one instant, the electron experiences an upward magnetic force of 1.6 × 10−13 N.

In what direction is the electron moving at that instant? [Hint: If there is more than one possible answer, find all the possibilities.]

Strategy.

This example is more complicated than Examples 19.1 and 19.2. We need to apply the magnetic force law again, but this time we must deduce the direction of the velocity from the directions of the force and field.

The direction of the magnetic force is up, so the direction of must be down since the charge is negative.

By the right-hand rule, the velocity must be somewhere in the left half of the plane; in other words, it must have a west component in addition to a north or south component.

The westward component is the component of v that is perpendicular to the field.

BF q B

1 6 10 N7 14 10 m/s

1 6 10 C 1 4 T

The velocity also has a component in the direction of the field that can be found using the Pythagorean theorem:

Use of the right-hand rule confirms that either gives in the correct direction.

2 2 61 87 10 m/s

Since 159° W of N is the same as 21° W of S, the direction of the velocity is either 21° W of N or 21° W of S.

7 14 10 m/ssin 0 357

2 0 10 m/s

21 W of N or159 W of N

19.3 Charged Particle Moving Perpendicularly to a Uniform Magnetic Field

Using the magnetic force law and Newton’s second law of motion, we can deduce the trajectory of a charged particle moving in a uniform magnetic field with no other forces acting.

In this section, we discuss a case of particular interest: when the particle is initially moving perpendicularly to the magnetic field.

Charged Particle Moving Perpendicularly to a Uniform Magnetic Field 2

Since v⊥ = v, the magnitude of the force is

Since the force is perpendicular to the velocity, the particle changes direction but not speed.

The force is also perpendicular to the field, so there is no acceleration component in the direction of Thus, the particle’s velocity remains perpendicular to

As the velocity changes direction, the magnetic force changes direction to stay perpendicular to both and The magnetic force acts as a steering force, curving the particle around in a trajectory of radius r at constant speed.

The particle undergoes uniform circular motion, so its acceleration is directed radially inward and has magnitude v2/r.

From Newton’s second law,

where m is the mass of the particle. Since the radius of the trajectory is constant— r depends only on q, v, B, and m, which are all constant—the particle moves in a circle at constant speed.

F q Ba

Negative charges move in the opposite sense from positive charges in the same field.

Application: Mass Spectrometer 1

Example 19.4

In a mass spectrometer, a beam of 6Li+ and 7Li+ ions passes through a velocity selector so that the ions all have the same velocity. The beam then enters a region of uniform magnetic field.

If the radius of the orbit of the 6Li+ ions is 8.4 cm, what is the radius of the orbit of the 7Li+ ions?

Example 19.4 Strategy

The charge of the 6Li+ ions is the same as the charge of the 7Li+ ions.

The ions enter the magnetic field with the same speed. We do not know the magnitudes of the charge, velocity, or magnetic field, but they are the same for the two types of ion.

With so many common quantities, a good strategy is to try to find the ratio between the radii for the two types of ions so that the common quantities cancel out.

Example 19.4 Solution

From Appendix B:6 6 015 um

7 7 016 um 271 u 1.66 10 kg

2 q BFa

7 016 u1 166

6 015 u

7 8 4 cm 1 166 9 8 cmr

Example 19.4 Comments

The example problem is actually the reverse of what is done. Ions are passed through the magnetic field, the radius of the orbit measured, and the molar mass calculated.

This is also a method to separate ions. Saddam Hussein used this method to separate 238UF6

+ from 235UF6+ to attempt to build an

illegal nuclear weapon.

If the magnetic field is about 2 T (not particularly hard to build) and the molecules have the same speed of about , calculate the radius of the orbit of the two molecules. (your answers should differ by about 0.1 mm!)

9⋅103m / s

Application: Cyclotrons

Example 19.5

A proton cyclotron uses a magnet that produces a 0.60-T field between its poles. The radius of the dees is 24 cm.

What is the maximum possible kinetic energy of the protons accelerated by this cyclotron?

Strategy.

As a proton’s kinetic energy increases, so does the radius of its path in the dees.

The maximum kinetic energy is therefore determined by the maximum radius.

Example 19.5 Solution2mv

F q Br

q BrK mv m

1 6 10 C 0 60 T 0 24 m

2 2 1 67 10 kg

1 6 10 J

19.4 Motion of a Charged Particle in a Uniform Magnetic Field: General

In general, the velocity may have components both perpendicular to and parallel to the magnetic field.

The component parallel to the field is constant, since the magnetic force is always perpendicular to the field.

The particle therefore moves along a helical path. The helix is formed by circular motion of the charge in a plane perpendicular to the field superimposed onto motion of the charge at constant speed along a field line.

Application : Aurorae on Earth, Jupiter, and Saturn

19.5 A Charged Particle in Crossed E and B Fields 1

For any particular combination of electric and magnetic fields, the two forces are balanced only for one particular particle speed, since the magnetic force is velocity-dependent, but the electric force is not. The velocity that gives zero net force can be found from:

There is zero net force on the particle only if

and if the direction of is correct. Since , it can be shown that the correct direction of is the direction of .

Since , it can be shown that the correct direction of is the direction of .

Recall that reversing the order of a cross product reverses the direction. So we know that:

and since v and B are perpendicular, E = Bv

(!!show that the units work).

Use the RHR to show that and

!! Why is it obviously wrong to say that:

E⃗=B⃗ x v⃗

v⃗∝ E⃗ x B⃗ B⃗∝ v⃗ x E⃗

v⃗=E⃗ x B⃗

Application: Velocity Selector

A velocity selector uses crossed electric and magnetic fields to select a single velocity out of a beam of charged particles.

Example 19.6

A velocity selector is to be constructed to select ions moving to the right at 6.0 km/s. The electric field is 300.0 V/m into the page.

What should be the magnitude and direction of the magnetic field?

First, in a velocity selector, E, B, and v are mutually perpendicular. That allows only two possibilities for the direction of B.

Setting the magnetic force equal and opposite to the electric force determines which of the two directions is correct and gives the magnitude of B.

The magnitude of the magnetic field is chosen so that the electric and magnetic forces on a particle moving at the given speed are exactly opposite.

The sign of the ions’ charge is irrelevant—changing the charge from positive to negative would change the directions of both forces, leaving them still opposite to each other.

For simplicity, then, we assume the charge to be positive.

Using the right-hand rule to evaluate both possibilities for B (up and down), we find that is out of the page if B is up.

The magnitudes of the forces must also be equal:

(Show that a is a and also a Tesla.)

qE q B

300 0 V/m0 050 T

6000 m/s

V⋅sm2

V /mm / s

Discovery of the Electron 1

The velocity selector can be used to determine the charge-to-mass ratio q /m of a charged particle.

First, the particle is accelerated from rest through a potential difference ΔV, converting electric potential energy into kinetic energy.

The change in its electric potential energy is ΔU = qΔV, so the charge acquires a kinetic energy.

2K mv q V

Now a velocity selector is used to determine the speed v = E/B, by adjusting the electric and magnetic fields until the particles pass straight through.

The charge-to-mass ratio q/m can now be determined.

If that number is a constant, then that is strong evidence that mass and charge must ALWAYS travel together, rather than being two properties that can be separated and treated independently.

This would indicate that charge and mass are two properties of ONE entity (or particle).

By measuring the charge-to-mass ratio, Thomson established that cathode rays are streams of negatively charged particles that all have the same charge-to-mass ratio—particles we now call electrons.

Application: Electromagnetic Blood Flowmeter

E BF F

qE q B

The Hall Effect 1

The Hall effect is similar in principle to the electromagnetic flowmeter, but pertains to the moving charges in a current-carrying wire or other solid, not to moving ions in blood.

A magnetic field perpendicular to the wire causes the moving charges to be deflected to one side. This charge separation causes an electric field across the wire.

The potential difference (or Hall voltage ) across the wire is measured and used to calculate the electric field (or Hall field ) across the wire.

The Hall Effect 2

The drift velocity of the charges is then given by vD = E/B. The Hall effect enables the measurement of the drift velocity and the determination of the sign of the charges.

(The carriers in metals are generally electrons, but semiconductors may have positive or negative carriers or both.)

The Hall effect is also the principle behind the Hall probe, a common device used to measure magnetic fields.

Example 19.7(1)

A flat slab of semiconductor has thickness t = 0.50 mm, width w = 1.0 cm, and length L = 30.0 cm. A current I = 2.0 A flows along its length to the right. A magnetic field B = 0.25 T is directed into the page, perpendicular to the flat surface of the slab.

Example 19.7(2)

Assume that the carriers are electrons. There are 7.0 × 1024

mobile electrons per m3.

(a) What is the magnitude of the Hall voltage across the slab?

(b) Which edge (top or bottom) is at the higher potential?

Given: current 2.0 A, magnetic field 0.25 T,

thickness 0.50 10 m, width 0.010 m,

7.0 10 electrons/ m

(a)DI neA

E H B DF eE F e B

H DE B

H H DV E w B w

(a) continued.

BIw BIV

newt net

24 3 19 3

0 25 T 2 0 A

7 0 10 m 1 6 10 C 0 50 10 m

0 89 mV

(b) Since the current flows to the right, the electrons actually move to the left.

(b) continued.

The bottom edge is at the higher potential.

19.6 Magnetic Force on a Current-Carrying Wire 1

A straight wire segment of length L in a uniform magnetic field B carries a current I. The mobile carriers have charge q. The magnetic force on any one charge is

where v is the instantaneous velocity of that charge.

The net magnetic force on the wire is the vector sum of these forces. Instead of summing the instantaneous magnetic force on each charge, we can instead multiply the average magnetic force on each charge by the number of charges.

Since each charge has the same average velocity—the drift velocity—each experiences the same average magnetic force:

DI nqA

number per unit volume volumeN nLA

D DNq nqAL

F v B v B

Almost there! Since current is not a vector, we cannot substitute.

Therefore, we define a length vector L to be a vector in the direction of the current with magnitude equal to the length of the wire.

and...

DnqAL I

Magnetic force on a straight segment of current-carrying wire

The magnitude of the force is

The direction of the force is perpendicular to both L and B.

sinF IL B ILB ILB

Finding the Magnetic Force on a Straight Segment of Current-Carrying Wire

1. The magnetic force is zero if (a) the current in the wire is zero, (b) the wire is parallel to the magnetic field, or (c) the magnetic field is zero.

2. Otherwise, determine the angle θ between L and B when the two are drawn starting at the same point.

3. Find the magnitude of the force from

4. Determine the direction of using the right-hand rule.

sinF IL B ILB ILB

Example 19.8

A 125-m-long power line is horizontal and carries a current of 2500 A toward the south. The Earth’s magnetic field at that location is 0.052 mT toward the north and inclined 62° below the horizontal.

What is the magnetic force on the power line? (Ignore any drooping of the wire; assume it’s straight.)

I = 2500 A;

has magnitude 125 m and direction south;

has magnitude 0.052 mT. It has a downward component and a northward component.

We find the cross product and then multiply by I.

The direction of the force is east.

(A wire in this application would have a weight around 90 N, so the magnetic force is smaller but NOT insignificant).

F IL B ILB 5sin 62 2500 A 125 m 5 2 10 T sin 62

19.7 Torque on a Current Loop

Consider a rectangular loop of wire carrying current I in a uniform magnetic field B.

The torque tends to make the loop rotate.Access the text alternative for these images

Torque on a Current Loop 2

The magnitude of the magnetic force on sides 2 and 4 is

The lever arm for each of the two forces is (1/2)b, so the torque due to each is

Then the total torque on the loop is τ = IabB. The area of the rectangular loop is A = ab, so

F ILB IaB

1 1magnitude of force lever arm

2 2F b IabB

If, instead of a single turn, there are N turns forming a coil, then the magnetic torque on the coil is

The equation holds for a planar loop or coil of any shape.

What if the field is not parallel to the plane of the coil?

For θ ≠ 90°, the magnetic forces on sides 1 and 3 are no longer zero, but they are equal and opposite and act along the same line of action, so they contribute neither to the net force nor to the net torque.

The magnetic forces on sides 2 and 4 are the same as before, but now the lever arms are smaller by a factor of sin θ: instead of (1/2)b, the lever arms are now (1/2)b sin θ.

There are two positions of rotational equilibrium, but they are not equivalent.

The position at θ = 180° is an unstable equilibrium because at angles near 180° the torque tends to rotate the coil away from 180°.

The position at θ = 0° is a stable equilibrium; the torque for angles near 0° makes the coil rotate back toward θ = 0° and thus tends to restore the equilibrium.

sin NIAB

Current Loops are Magnetic Dipoles

The torque on a current loop in a uniform magnetic field is analogous to the torque on an electric dipole in a uniform electric field.

This similarity is our first hint that

A current loop is a magnetic dipole.

The direction perpendicular to the loop chosen by the right-hand rule is the direction of the magnetic dipole moment vector. The dipole moment vector points from the dipole’s south pole toward its north pole.

Application: Electric Motor 1

Application: Galvanometer

Application: Audio Speakers

19.8 Magnetic Field due to an Electric Current 1

We have not yet looked at sources of magnetic fields other than permanent magnets. It turns out that any moving charged particle creates a magnetic field. There is a certain symmetry about the situation:

• Moving charges experience magnetic forces and moving charges create magnetic fields;

• Charges at rest feel no magnetic forces and create no magnetic fields;

• Charges feel electric forces and create electric fields, whether moving or not.

The magnetic field due to a single moving charged particle is negligibly small in most situations.

However, when an electric current flows in a wire, there are enormous numbers of moving charges.

The magnetic field due to the wire is the sum of the magnetic fields due to each charge; the principle of superposition applies to magnetic fields just as it does to electric fields.

Magnetic Field due to a Long Straight Wire 1

©Science Source; ©GIPhotoStock/Science Source

Using a Right-Hand Rule to Find the Direction of the Magnetic Field due to a Long Straight Wire

1. Point the thumb of the right hand in the direction of the current in the wire.

2. Curl the fingers inward toward the palm; the direction that the fingers curl is the direction of the magnetic field lines around the wire.

3. As always, the magnetic field at any point is tangent to a field line through that point. For a long straight wire, the magnetic field is tangent to a circular field line and, therefore, perpendicular to a radial line from the wire.

The magnitude of the magnetic field at a distance r from the wire can be found using Ampère’s law (Section 19.9; see Example 19.11):

where I is the current in the wire and 0 is a universal constant known as the permeability of free space.

The permeability plays a role in magnetism similar to the role of the permittivity (ϵ0) in electricity.

Permeability of Free Space

In SI units, the value of 0 is

The constant 0 can be assigned an exact value because the magnetic forces on two parallel wires are used to define the ampere, which is an SI base unit.

One ampere is the current in each of two long parallel wires 1 m apart such that each exerts a magnetic force on the other of exactly 2 × 10−7 N per meter of length.

T m4 10 exact by definition

Forces on Parallel Current-Carrying Wires 1

Two parallel current-carrying wires that are close together exert magnetic forces on one another.

The magnetic field of wire 1 causes a magnetic force on wire 2; the magnetic field of wire 2 causes a magnetic force on wire 1.

From Newton’s third law, we expect the forces on the wires to be equal and opposite.

If the currents flow in the same direction, the force is attractive; if they flow in opposite directions, the force is repulsive.

Example 19.10

In household wiring, two long parallel wires are separated and surrounded by an insulator. The wires are a distance d apart and carry currents of magnitude I in opposite directions.

(a) Find the magnetic field at a distance r >> d from the center of the wires (point P in the figure).

(b) Find the numerical value of B if I = 5 A, d = 5 mm, and r = 1 m and compare with Earth’s magnetic field strength at the surface (≈ 5 × 10−5 T).

The magnetic field is the vector sum of the fields due to each of the wires.

The fields due to the wires at P are equal in magnitude (since the currents and distances are the same), but the directions are not the same. The equation.

gives the magnitude of the field due to either wire. Since the field lines due to a single long wire are circular, the direction of the field is tangent to a circle that passes through P and whose center is on the wire.

The right-hand rule determines which of the two tangent directions is correct.

IB r d

0 sin2x

(a) continued.

The field due to the wires is 10−4 times Earth’s field.

1opposite 2sin

hypotenuse

7 902 2

T m 5 A 0.005 m2 10 5 10 T

2 A (1 m)

Bnet=2B x=μ0 I d

2π r2

Use the same expression and consider a high voltage power line. The current is about 1500 A and the wires are about 2 m apart. How far would you need to be for the magnetic field to be equal to the strength of the Earth’s magnetic field?

Start with:

And solve for r:

(I get about 3.5 m)

Bnet=2B x=μ0 I d

2π r2

Magnetic Field due to a Circular Current Loop

In Section 19.7, we saw the first clue that a loop of wire that carries current around in a complete circuit is a magnetic dipole.

A second clue comes from the magnetic field produced by a circular loop of current.

Using a Right-Hand Rule to Find the Direction of the Magnetic Field due to a Loop of Current

Curl the fingers of your right hand inward toward the palm, following the current around the loop.

Your thumb points in the direction of the magnetic field in the interior of the loop.

Magnetic Field at the Center of a Circular Loop of Current

The magnitude of the magnetic field at the center of a circular loop (or coil) is given by.

where N is the number of turns, I is the current, and r is the radius.

Magnetic Field due to a Solenoid 1

An important source of magnetic field is that due to a solenoid because the field inside a solenoid is nearly uniform.

In magnetic resonance imaging (MRI), the patient is immersed in a strong magnetic field inside a solenoid.

Recall the permanent magnet field from Section 19.1

Magnetic Field of a Solenoid

Comparing the two suggests that the permanent magnet field is caused by currents in the atoms making up the magnet. (These currents create magnetic fields, but we can’t seem to make them light a bulb :)

If a long solenoid has N turns of wire and length L, then the magnetic field strength inside is given by:

Magnetic field strength inside an ideal solenoid:

In the equation, I is the current in the wire and n = N/L is the number of turns per unit length.

Note that the field does not depend on the radius of the solenoid.

NIB nI

Application: Magnetic Resonance Imaging

19.9 Ampère’s Law 1

Ampère’s law plays a role in magnetism similar to that of Gauss’s law in electricity.

Both relate the field to the source of the field. For the electric field, the source is charge. The source of magnetic fields is current.

Ampère’s law must take a different form from Gauss’s law: since magnetic field lines are always closed loops, the magnetic flux through a closed surface is always zero. (This fact is called Gauss’s law for magnetism and is itself a fundamental law of electromagnetism.)

Instead of a closed surface, Ampère’s law concerns any closed path or loop.

For Gauss’s law we would find the flux: the perpendicular component of the electric field times the surface area.

For Ampère’s law, we multiply the component of the magnetic field parallel to the path (or the tangential component at points along a closed curve) times the length of the path.

Circulation

Just as for flux, if the magnetic field component is not constant then we take parts of the path (each of length Δl ) and sum up the product.

This quantity is called the circulation.

||circulation B l

Comparison of Gauss’s and Ampère’s Laws

Table 19.1 Comparison of Gauss’s and Ampère’s Laws

Gauss’s Law Ampère’s Law

Electric field Magnetic field (static only)

Applies to any closed surface Applies to any closed path

Relates the electric field on the surface tothe net electric charge inside the surface

Relates the magnetic field on the path to the net current cutting through interior of the path

Component of the electric fieldperpendicular to the surface (E⊥)

Component of the magnetic field parallelto the path (B||)

Flux = perpendicular field component ×area of surface

Circulation = parallel field component ×length of path

Flux = 1/ϵ0 × net charge Circulation = μ0 × net currentE A

1E A q

|| 0B l I

Ampère’s law relates the circulation of the field to the net current I that crosses the interior of the path.

Ampère’s Law.

|| 0B l I

Example 19.11

Use Ampère’s law to show that the magnetic field due to a long straight wire is B = 0I/(2r).

As with Gauss’s law, the key is to exploit the symmetry of the situation. The field lines have to be circles around the wire, assuming the ends are far away.

Choose a closed path around a circular field line. The field is everywhere tangent to the field line and therefore tangent to the path; there is no perpendicular component.

The field must also have the same magnitude at a uniform distance r from the wire.

Since the field has no component perpendicular to the path, B|| = B. Going around the circular path, B is constant, so

0Circulation 2B r I

Giambattista College Physics Chapter 19

©2020 McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education.

Chapter 19: Magnetic Forces and Fields

19.1 Magnetic Fields.

19.2 Magnetic Force on a Point Charge.

19.3 Charged Particle Moving Perpendicularly to a Uniform Magnetic Field.

19.4 Motion of a Charged Particle in a Uniform Magnetic Field: General.

19.5 A Charged Particle in Crossed E and B Fields.

19.6 Magnetic Force on a Current-Carrying Wire.

19.7 Torque on a Current Loop.

19.8 Magnetic Field due to an Electric Current.

19.9 Ampère’s Law.

19.10 Magnetic Materials.

A bar magnet is one instance of a magnetic dipole. By dipole we mean two opposite poles.

Look carefully and note that themagnetic field lines do NOT begin and end like electricfield lines, but form closed loops.

They SEEM to begin at the N pole andend at the S pole, but they really reconnect inside. This was not obviousin the beginning of the study of magnets (1400s!) but will be suggested by the magnetic field of a solenoid (Ch 19.8).

EDITORIAL COMMENT:

I’m generally very unhappy with the way that magnetic field lines are drawn, since studentswill USUALLY have to UNLEARN what these diagramsteach. These aren’t that bad (we’ll seeworse examples).

But we will look at this in lab (if you havelab with me) and I’ll try to stress the difference between reality and the imagination of people who draw these diagrams.

Permanent Magnets 1

In an electric dipole, the electric poles are positive and negative electric charges. A magnetic dipole consists of two opposite magnetic poles.

The end of the bar magnet where the field lines emerge is called the north pole, and the end where the lines go back in is called the south pole. Like poles repel one another and opposite poles attract one another.

Ultimately, magnetic fields and forces are created and felt by currents – moving charges.

The magnetic field is represented by the symbol

Permanent Magnets 2

A compass is simply a small bar magnet that is free to rotate.

Any magnetic dipole, including a compass needle, feels a torque that tends to line it up with an external magnetic field.

A magnet need not have only two poles; it must have at least one north pole and at least one south pole.

Permanent Magnets 2

A compass is simply a small bar magnet that is free to rotate, but NOT move.

Any magnetic dipole, including a compass needle, feels a TORQUE that tends to line it up with an external magnetic field.

There is also a net FORCE toward the area of higher magnetic field.

If you divide a magnet in half, you get TWO magnets, each with a N and S pole.

Ultimately, if you keep going, you will have a single atom with a N and a S pole.

Ultimately, we will see that these fields are created by loops of current, with a N pole on one side of the current, and a S pole on the other.

NOT by magnetic particles.

Figure used w permission, from:http://www.gonefcon.com/trucktcom/magnets.htm

Does a good job showing what happens when the magnet is divided – the rest, not so good.

No Magnetic Monopoles

Coulomb’s law for electric forces gives the force acting between two point charges—two electric monopoles.

However, as far as we know, there are no magnetic monopoles—that is, there is no such thing as an isolated north pole or an isolated south pole.

There’s no REASON there shouldn’t exist, and some scientists have speculated that they SHOULD. So they are looking for them and will no doubt get a Nobel if they find one.

But science doesn’t care if they SHOULD or SHOULDN’T. Whether they exist is an EMPIRICAL question – and so far, no one’s ever found one.

Magnetic field lines do not begin on north poles and end on south poles: magnetic field lines are always closed loops.

If there are no magnetic monopoles, there is no place for the field lines to begin or end, so they must be closed loops.

(Later, we’ll see that they loop AROUND the electric currents that CREATE the field).

Interpretation of Magnetic Field Lines

Same as the interpretation of electric field lines.

The direction of the magnetic field vector at any point is tangent to the field line passing through that point and is in the direction indicated by arrows on the field line.

The magnetic field is strong where field lines are close together and weak where they are far apart.

More specifically, if you imagine a small surface perpendicular to the field lines, the magnitude of the magnetic field is proportional to the number of lines that cross the surface, divided by the area.

(If that sounds like “flux”, it’s not an accident.)

The Earth’s Magnetic Field

The south pole of the fictitious magnet faces roughly toward geographic north and the north pole of the magnet faces roughly toward geographic south. (Field lines could be better)

In Chapter 16 we defined the electric field as the electric force per unit charge. The electric force is either in the same direction as the electric field or in the opposite direction, depending on the sign of the point charge.

The magnetic force on a point charge is more complicated—it is not the charge times the magnetic field. The magnetic force depends on the point charge’s velocity as well as on the magnetic field.

If the point charge is at rest, there is no magnetic force.

The magnitude and direction of the magnetic force depend on the direction and speed of the charge’s motion.

We have learned about other velocity-dependent forces, such as the drag force on an object moving through a fluid. Like drag forces, the magnetic force increases in magnitude with increasing velocity.

However, the direction of the drag force is always opposite to the object’s velocity, while the direction of the magnetic force on a charged particle is perpendicular to the velocity of the particle.

MAGNITUDE: We defined an electric field to model the force that two charges exert upon one another. If those charges are moving, there is ANOTHER force that behaves differently:The magnitude of the force is proportional to the speed, the charge (incl sign) and the strength of the field. So the force is proportional to the product:

BUT, if the velocity is parallel to the magnetic field, the force is zero AND is a maximum if the velocity is perpendicular to the magnetic field. A function that is zero when and maximum when is the sine function. So, we can summarize that the magnitude of the magnetic force on a moving charge is:

F ~qvB

θ =0θ =90°=π /2

F=qvB sin(θ )

UNITS: W give the unit for the magnetic field it’s own name (unlike the electric field): The units of magnetic field are Tesla (named for the scientist, not the car). Starting with our expression for the force, we can put all the units in asn see how to write a Tesla in terms of other units:

So a Tesla can be written as:

We’ll use this often to keep track of units when we calculate the magnetic field of various current arrangements.

F=qvB sin(θ )[N ]=[C ][m /s ][T ][ ]