Physics1C Chapter 27 Slides

Physics 1C Chapter 27

Magnetic Field and Magnetic Forces

Vahé Peroomian

Sections Covered •  Covered Sections:

Sections 27.1 – 27.7

•  Important Sections: Sections 27.2 – 27.4, 27.6 – 27.7

•  Sections Skipped: Sections 27.8 – 27.9 + magnetic dipole in a non-

uniform magnetic field, magnetic dipoles and how magnets work.

You will not be responsible for sections I skip on either homework assignments or on exams.

Magnetism

•  Magnets always have a north pole and a south pole.

•  Just as in electric charges, like poles repel and opposite poles attract.

Magnetism

•  If you cut a bar magnet in half, you get two magnets both with north and south poles.

•  Some materials, called ferromagnets, are naturally magnetic.

Magnetic Fields

•  The direction of the magnetic field vector B at any location is the direction in which the north pole of a compass needle points at that location.

•  Magnetic field lines begin at the north pole and end at the south pole.

•  The existence of a magnetic field at some point in space can be determined by measuring the magnetic force exerted on a charged test particle placed at that point.

Earth’s Magnetic Field

Magnetic Forces and Fields

•  Similarities between magnetic and electric fields: o The magnetic force is proportional to the charge q of

the particle. o The magnetic force on a negative charge is directed

opposite to the force on a positive charge moving in the same direction.

o The magnetic force is proportional to the magnitude of the magnetic field vector B.


•  Differences between magnetic and electric fields: o  The magnetic force is proportional to the speed v of the

particle. o  If the velocity vector makes an angle θ with the magnetic field,

the magnitude of the magnetic force is proportional to sinθ. o  When a charged particle moves parallel to the magnetic field

vector, the magnetic force on the charge is zero. o  When a charged particle moves in a direction not parallel to the

magnetic field vector, the magnetic force acts in a direction perpendicular to both v and B; that is, the magnetic force is perpendicular to the plane formed by v and B.


•  The magnetic force on a charged particle moving in a magnetic field is

•  The SI unit of magnetic field is the tesla (T): 1 T = 1 N . S / C . m

FB = q

v ×B

746$$CHAPTER 22 | Magnetic Forces and Magnetic Fields

Figure 22.4 Two right-hand rules for determining the direction of the magnetic force F

:B 5 qv: 3 B

:

acting on a particle with positive charge q moving with a velocity v: in a magnetic fi eld B

:. (a) In this rule,

the magnetic force is in the direction in which your thumb points. (b) In this rule, the magnetic force is in the direction of your palm, as if you are pushing the particle with your hand.

BS

FBS

FBS

a b

(1) Point your fingers in the direction of v and then curl them toward the direction of B.

S

S

(1) Point your fingers in the direction of B, with v coming out of your thumb.

S

S

BS

vS

vS

(2) Your upright thumb shows the direction of the magnetic force on a positive particle.

(2) The magnetic force on a positive particle is in the direction you would push with your palm.

velocity of the particle and because its direction is perpendicular to both v: and B:

. Figure 22.3 shows the details of the direction of the magnetic force on a charged particle. Despite this complicated behavior, these observations can be summarized in a compact way by writing the magnetic force in the form

F:

B 5 q v: 3 B:

22.1b

where the direction of the magnetic force is that of v: 3 B:

, which, by defi nition of the cross product, is perpendicular to both v: and B

:. Equation 22.1 is analogous to

Equation 19.4, F:

e 5 qE:

, but is clearly more complicated. We can regard Equa-tion 22.1 as an operational defi nition of the magnetic fi eld at a point in space. The SI unit of magnetic fi eld is the tesla (T), where

1 T 5 1 N ? s/C ? m

Figure 22.4 reviews two right-hand rules for determining the direction of the cross product v: 3 B

: and determining the direction of F

:B. The rule in Figure 22.4a depends

on our right-hand rule for the cross product in Figure 10.13. You point the four fi n-gers of your right hand along the direction of v: with the palm facing B

: and curl them

toward B:

. The extended thumb, which is at a right angle to the fi ngers, points in the direction of v: 3 B

:. Because F

:B 5 qv: 3 B

:, F

:B is in the direction of your thumb if q is

positive and opposite the direction of your thumb if q is negative. A second rule is shown in Figure 22.4b. Here the thumb points in the direction

of v: and the extended fi ngers in the direction of B:

. Now, the force F:

B on a positive charge extends outward from your palm. The advantage of this rule is that the force on the charge is in the direction that you would push on something with your hand, outward from your palm. The force on a negative charge is in the opposite direction. Feel free to use either of these two right-hand rules.

c Vector expression for the magnetic force on a charged particle moving in a magnetic fi eld

Figure 22.3 (a) The direction of the magnetic force F

:B acting

on a charged particle moving with a velocity v: in the presence of a magnetic fi eld B

:. (b) Magnetic forces

on positive and negative charges. The dashed lines show the paths of the particles, which are investigated in Section 22.3.

u

vS

vS

vS

FBS

FBS

FBS

BS

BS

!

!

"

The magnetic forces on oppositely charged particles moving at the same velocity in a magnetic field are in opposite directions.

a b

The magnetic force is perpendicular to both v and B.S S

04261_ch22_741-780.indd 74604261_ch22_741-780.indd 746 24/11/11 2:25 PM24/11/11 2:25 PM


Figure 22.4 Two right-hand rules for determining the direction of the magnetic force F

:B 5 qv: 3 B

:

acting on a particle with positive charge q moving with a velocity v: in a magnetic fi eld B

:. (a) In this rule,

the magnetic force is in the direction in which your thumb points. (b) In this rule, the magnetic force is in the direction of your palm, as if you are pushing the particle with your hand.

BS

FBS

FBS

a b

(1) Point your fingers in the direction of v and then curl them toward the direction of B.

S

S

(1) Point your fingers in the direction of B, with v coming out of your thumb.

S

S

BS

vS

vS

(2) Your upright thumb shows the direction of the magnetic force on a positive particle.

(2) The magnetic force on a positive particle is in the direction you would push with your palm.

velocity of the particle and because its direction is perpendicular to both v: and B:

. Figure 22.3 shows the details of the direction of the magnetic force on a charged particle. Despite this complicated behavior, these observations can be summarized in a compact way by writing the magnetic force in the form

F:

B 5 q v: 3 B:

22.1b

where the direction of the magnetic force is that of v: 3 B:

, which, by defi nition of the cross product, is perpendicular to both v: and B

:. Equation 22.1 is analogous to

Equation 19.4, F:

e 5 qE:

, but is clearly more complicated. We can regard Equa-tion 22.1 as an operational defi nition of the magnetic fi eld at a point in space. The SI unit of magnetic fi eld is the tesla (T), where

1 T 5 1 N ? s/C ? m

Figure 22.4 reviews two right-hand rules for determining the direction of the cross product v: 3 B

: and determining the direction of F

:B. The rule in Figure 22.4a depends

on our right-hand rule for the cross product in Figure 10.13. You point the four fi n-gers of your right hand along the direction of v: with the palm facing B

: and curl them

toward B:

. The extended thumb, which is at a right angle to the fi ngers, points in the direction of v: 3 B

:. Because F

:B 5 qv: 3 B

:, F

:B is in the direction of your thumb if q is

positive and opposite the direction of your thumb if q is negative. A second rule is shown in Figure 22.4b. Here the thumb points in the direction

of v: and the extended fi ngers in the direction of B:

. Now, the force F:

B on a positive charge extends outward from your palm. The advantage of this rule is that the force on the charge is in the direction that you would push on something with your hand, outward from your palm. The force on a negative charge is in the opposite direction. Feel free to use either of these two right-hand rules.

c Vector expression for the magnetic force on a charged particle moving in a magnetic fi eld

Figure 22.3 (a) The direction of the magnetic force F

:B acting

on a charged particle moving with a velocity v: in the presence of a magnetic fi eld B

:. (b) Magnetic forces

on positive and negative charges. The dashed lines show the paths of the particles, which are investigated in Section 22.3.

u

vS

vS

vS

FBS

FBS

FBS

BS

BS

!

!

"

The magnetic forces on oppositely charged particles moving at the same velocity in a magnetic field are in opposite directions.

a b

The magnetic force is perpendicular to both v and B.S S



•  We find the direction of the magnetic force on a charged particle by using the right-hand rule.

•  The magnitude of the magnetic force on a charged particle is

FB = q vBsinθ

Difference Between Magnetic and Electric Forces •  Important differences between electric and

magnetic forces on charged particles: o  The electric force vector is along the direction of the

electric field, whereas the magnetic force vector is perpendicular to the magnetic field.

o  The electric force acts on a charged particle regardless of whether the particle is moving, whereas the magnetic force acts on a charged particle only when the particle is in motion.

o  The electric force does work in displacing a charged particle, whereas the magnetic force associated with a steady magnetic field does no work when a particle is displaced.

How We Draw Magnetic Fields •  Field out of the page:

•  Field into the page:

22.2 | The Magnetic Field""747

The magnitude of the magnetic force is

FB 5 u q uvB sin ! 22.2b

where ! is the angle between v: and B:

. From this expression, we see that FB is zero when v: is either parallel or antiparallel to B

: (! 5 0 or 1808). Furthermore, the force

has its maximum value FB 5 u q uvB when v: is perpendicular to B:

(! 5 908).Let’s summarize the important differences between electric and magnetic forces

on charged particles:

• The electric force vector is along the direction of the electric fi eld, whereas the magnetic force vector is perpendicular to the magnetic fi eld.

• The electric force acts on a charged particle regardless of whether the particle is moving, whereas the magnetic force acts on a charged particle only when the particle is in motion.

• The electric force does work in displacing a charged particle, whereas the mag-netic force associated with a steady magnetic fi eld does no work when a particle is displaced.

This last statement is true because when a charge moves in a constant magnetic fi eld, the magnetic force is always perpendicular to the displacement of its point of application. That is, for a small displacement d s: of a particle, the work done by the magnetic force on the particle is dW 5 F

:B ? d s: 5 (F

:B ? v:)dt 5 0 because the mag-

netic force is a vector perpendicular to v:. From this property and the work–kinetic energy theorem, we conclude that the kinetic energy of a charged particle cannot be altered by a constant magnetic fi eld alone. In other words, when a charge moves with a velocity of v:, an applied magnetic fi eld can alter the direction of the velocity vector, but it cannot change the speed of the particle.

In Figures 22.3 and 22.4, we used green arrows to represent magnetic fi eld vec-tors, which will be the convention in this book. In Active Figure 22.1, we represented the magnetic fi eld of a bar magnet with green fi eld lines. Studying magnetic fi elds presents a complication that we avoided in electric fi elds. In our study of electric fi elds, we drew all electric fi eld vectors in the plane of the page or used perspective to represent them directed at an angle to the page. The cross product in Equation 22.1 requires us to think in three dimensions for problems in magnetism. Therefore, in addition to drawing vectors pointing left or right and up or down, we will need a method of drawing vectors into or out of the page. These methods of representing the vectors are illustrated in Figure 22.5. A vector coming out of the page is repre-sented by a dot, which we can think of as the tip of the arrowhead representing the vector coming through the paper toward us (Fig. 22.5a). A vector going into the page is represented by a cross, which we can think of as the tail feathers of an arrow going into the page (Fig. 22.5b). This depiction can be used for any type of vector we will encounter: magnetic fi eld, velocity, force, and so on.

QUICK QUIZ 22.1 An electron moves in the plane of this paper toward the top of the page. A magnetic fi eld is also in the plane of the page and directed toward the right. What is the direction of the magnetic force on the electron? (a) toward the top of the page (b) toward the bottom of the page (c) toward the left edge of the page (d) toward the right edge of the page (e) upward out of the page (f) downward into the page

c Magnitude of the magnetic force on a charged particle moving in a magnetic fi eld

Figure 22.5 Representations of magnetic fi eld lines perpendicular to the page.

a

b

Magnetic field lines coming out of the paper are indicated by dots, representing the tips of arrows coming outward.

Magnetic field lines going into the paper are indicated by crosses, representing the feathers of arrows going inward.

BoutS

BinS

THINKING PHYSICS 22.1

On a business trip to Australia, you take along your U.S.-made compass that you used in your Boy Scout days. Does this compass work correctly in Australia?

Reasoning Using the compass in Australia presents no problem. The north pole of the magnet in the com-pass will be attracted to the south magnetic pole near the north geographic pole, just as it was in the United

States. The only difference in the magnetic fi eld lines is that they have an upward component in Australia, whereas they have a downward component in the United States. When you hold the compass in a hori-zontal plane, it cannot detect the vertical component of the fi eld, however; it only displays the direction of the horizontal component of the magnetic fi eld. b


22.2 | The Magnetic Field""747

The magnitude of the magnetic force is

FB 5 u q uvB sin ! 22.2b

where ! is the angle between v: and B:

. From this expression, we see that FB is zero when v: is either parallel or antiparallel to B

: (! 5 0 or 1808). Furthermore, the force

has its maximum value FB 5 u q uvB when v: is perpendicular to B:

(! 5 908).Let’s summarize the important differences between electric and magnetic forces

on charged particles:

• The electric force vector is along the direction of the electric fi eld, whereas the magnetic force vector is perpendicular to the magnetic fi eld.

• The electric force acts on a charged particle regardless of whether the particle is moving, whereas the magnetic force acts on a charged particle only when the particle is in motion.

• The electric force does work in displacing a charged particle, whereas the mag-netic force associated with a steady magnetic fi eld does no work when a particle is displaced.

This last statement is true because when a charge moves in a constant magnetic fi eld, the magnetic force is always perpendicular to the displacement of its point of application. That is, for a small displacement d s: of a particle, the work done by the magnetic force on the particle is dW 5 F

:B ? d s: 5 (F

:B ? v:)dt 5 0 because the mag-

netic force is a vector perpendicular to v:. From this property and the work–kinetic energy theorem, we conclude that the kinetic energy of a charged particle cannot be altered by a constant magnetic fi eld alone. In other words, when a charge moves with a velocity of v:, an applied magnetic fi eld can alter the direction of the velocity vector, but it cannot change the speed of the particle.

In Figures 22.3 and 22.4, we used green arrows to represent magnetic fi eld vec-tors, which will be the convention in this book. In Active Figure 22.1, we represented the magnetic fi eld of a bar magnet with green fi eld lines. Studying magnetic fi elds presents a complication that we avoided in electric fi elds. In our study of electric fi elds, we drew all electric fi eld vectors in the plane of the page or used perspective to represent them directed at an angle to the page. The cross product in Equation 22.1 requires us to think in three dimensions for problems in magnetism. Therefore, in addition to drawing vectors pointing left or right and up or down, we will need a method of drawing vectors into or out of the page. These methods of representing the vectors are illustrated in Figure 22.5. A vector coming out of the page is repre-sented by a dot, which we can think of as the tip of the arrowhead representing the vector coming through the paper toward us (Fig. 22.5a). A vector going into the page is represented by a cross, which we can think of as the tail feathers of an arrow going into the page (Fig. 22.5b). This depiction can be used for any type of vector we will encounter: magnetic fi eld, velocity, force, and so on.

QUICK QUIZ 22.1 An electron moves in the plane of this paper toward the top of the page. A magnetic fi eld is also in the plane of the page and directed toward the right. What is the direction of the magnetic force on the electron? (a) toward the top of the page (b) toward the bottom of the page (c) toward the left edge of the page (d) toward the right edge of the page (e) upward out of the page (f) downward into the page

c Magnitude of the magnetic force on a charged particle moving in a magnetic fi eld

Figure 22.5 Representations of magnetic fi eld lines perpendicular to the page.

a

b

Magnetic field lines coming out of the paper are indicated by dots, representing the tips of arrows coming outward.

Magnetic field lines going into the paper are indicated by crosses, representing the feathers of arrows going inward.

BoutS

BinS

THINKING PHYSICS 22.1

On a business trip to Australia, you take along your U.S.-made compass that you used in your Boy Scout days. Does this compass work correctly in Australia?

Reasoning Using the compass in Australia presents no problem. The north pole of the magnet in the com-pass will be attracted to the south magnetic pole near the north geographic pole, just as it was in the United

States. The only difference in the magnetic fi eld lines is that they have an upward component in Australia, whereas they have a downward component in the United States. When you hold the compass in a hori-zontal plane, it cannot detect the vertical component of the fi eld, however; it only displays the direction of the horizontal component of the magnetic fi eld. b


Example 1 (from Serway & JeweI)

An electron in an old-style television picture tube moves toward the front of the tube with a speed of 8.0 × 106 m/s along the x axis. Surrounding the neck of the tube are coils of wire that create a magnetic field of magnitude 0.025 T, directed at an angle of 60° to the x axis and lying in the xy plane. Calculate the magnetic force on the electron.


22.3 | Motion of a Charged Particle in a Uniform Magnetic Field

In Section 22.2, we found that the magnetic force acting on a charged particle mov-ing in a magnetic fi eld is perpendicular to the particle’s velocity and consequently the work done by the magnetic force on the particle is zero. Now consider the spe-cial case of a positively charged particle moving in a uniform magnetic fi eld with the initial velocity vector of the particle perpendicular to the fi eld. Let’s assume the direction of the magnetic fi eld is into the page as in Active Figure 22.7. As the particle changes the direction of its velocity in response to the magnetic force, the magnetic force remains perpendicular to the velocity. As we found in Section 5.2, if the force is always perpendicular to the velocity, the path of the particle is a circle! Active Figure 22.7 shows the particle moving in a circle in a plane perpendicular to the magnetic fi eld. Although magnetism and magnetic forces may be new and unfa-miliar to you now, we see a magnetic effect that results in something with which we are familiar: the particle in uniform circular motion!

The particle moves in a circle because the magnetic force F:

B is perpendicular to v: and B

: and has a constant magnitude qvB. As Active Figure 22.7 illustrates, the

rotation is counterclockwise for a positive charge in a magnetic fi eld directed into the page. If q were negative, the rotation would be clockwise. We use the particle under a net force model to write Newton’s second law for the particle:

S F 5 FB 5 ma

Because the particle moves in a circle, we also model it as a particle in uniform circu-lar motion and we replace the acceleration with centripetal acceleration:

FB 5 qvB 5mv2

r

This expression leads to the following equation for the radius of the circular path:

r 5mvqB

22.3b

r

q

q

q

vS

vS

vS

FBS

FBS

FBS

BinS

!

!

!

The magnetic force FB acting on the charge is always directed toward the center of the circle.

S

Active Figure 22.7 When the ve-locity of a charged particle is perpen-dicular to a uniform magnetic fi eld, the particle moves in a circular path in a plane perpendicular to B

:.

Example 22.1 | An Electron Moving in a Magnetic Field

An electron in an old-style television picture tube moves toward the front of the tube with a speed of 8.0 3 106 m/s along the x axis (Fig. 22.6). Surrounding the neck of the tube are coils of wire that create a magnetic fi eld of magnitude 0.025 T, directed at an angle of 608 to the x axis and lying in the xy plane. Calculate the magnetic force on the electron.

SOLUTIONConceptualize Recall that the magnetic force on a charged particle is perpendicular to the plane formed by the velocity and magnetic fi eld vectors. Use one of the right-hand rules in Figure 22.4 to convince yourself that the direction of the force on the electron is downward in Figure 22.6.

Categorize We evaluate the magnetic force using an equation developed in this section, so we categorize this example as a substitution problem.

z

x

60"y

BS

FBS

vS

#e

Figure 22.6 (Example 22.1) The magnetic force F

:B

acting on the electron is in the negative z direction when v: and B

: lie in the xy plane.

Use Equation 22.2 to fi nd the magnitude of the magnetic force:

FB 5 |q |vB sin !

5 (1.6 3 10219 C)(8.0 3 106 m/s)(0.025 T)(sin 608)

5 2.8 3 10214 N

For practice using the vector product, evaluate this force in vector notation using Equation 22.1.


Motion of a Charged Particle in a Uniform Magnetic Field

•  Charged particles in a uniform magnetic field execute uniform circular motion with radius:

r = mv

qB

Motion of a Charged Particle in a Uniform Magnetic Field

•  The angular speed of the particle is

•  The period of the motion is

ω = v

r= qB

m

T = 2π

ω= 2πm

qB

Example 2 (from Serway & JeweI) In an experiment designed to measure the magnitude of a uniform magnetic field, electrons are accelerated from rest through a potential difference of 350 V and then enter a uniform magnetic field that is perpendicular to the velocity vector of the electrons. The electrons travel along a curved path because of the magnetic force exerted on them, and the radius of the path is measured to be 7.5 cm. (a) What is the magnitude of the magnetic field? (b) What is the angular speed of the electrons?

The LorenN Force

•  The total force on a charged particle moving in a region with both an electric and a magnetic field is

F = q

E + q

v ×B

In a velocity selector (see figure), the magnitudes of the electric and magnetic fields are chosen so that qE = qvB, the charged particle experiences zero net force and moves in a straight vertical line through the region of the fields. A certain velocity selector consists of electric and magnetic fields described by the expressions E = Ek and B = Bj with B = 15.0 mT. Find the value of E such that a 750-eV electron moving in the negative x direction is undeflected.


The Mass Spectrometer

•  Separates particles according to their mass/charge ratio.

•  Used on many spacecraft measuring the radiation environment in Earth orbit.

The Mass Spectrometer: Thompson’s Experiment

Particle Motion in Earth’s Magnetic Field

•  A magnetic mirror force acts on particles that move from smaller to larger magnetic field, reflecting them back along the magnetic field line.

•  The larger the pitch angle of the particle (the angle between the particle’s velocity vector and the magnetic field), the more effective this force.

•  Small pitch angle particles sneak through…

The Aurora Borealis

•  aa

Magnetic Flux

•  Magnetic flux through a surface is defined as

ΦB =B ⋅dA∫

Magnetic Flux

Gauss’s Law for Magnetism

•  Since there are no magnetic monopoles, the total magnetic flux through a closed surface is always zero.

ΦB =B ⋅dA∫ = 0

A rectangular loop of width a and length b is located near a long wire carrying current I. The distance between the wire and the closest side of the loop is c. The wire is parallel to the long side of the loop. Find the total magnetic flux through the loop due to the current in the wire.


Magnetic Force on a Current-‐‑Carrying Conductor

•  The magnetic force on the wire is due to the sum of the individual magnetic forces on the charged particles.

Magnetic Force on a Current-‐‑Carrying Conductor

•  The magnetic force on a straight wire is given by

•  For a wire of arbitrary shape,

FB = I

×B

dFB = Id

×B

FB = I d

×B

a

b

∫


A wire having a mass per unit length of 0.500 g/cm carries a 2.00-A current horizontally to the south. What are (a)  the direction and (b)  the magnitude of the minimum magnetic field

needed to lift this wire vertically upward?

Example 6 (from Giancoli) A 2.0-m-long wire carries a current of 8.2 A and is immersed in a uniform magnetic field B. When this wire lies along the +x axis, a magnetic force F =(–2.5 N)j acts on the wire, and when it lies on the +y axis, the force is F = (2.5 N)i – (5.0 N)k. Find B.

Example 7 (from Serway & JeweI) A wire bent into a semicircle of radius R forms a closed circuit and carries a current I. The wire lies in the xy plane, and a uniform magnetic field is directed along the positive y axis as in the figure. Find the magnitude and direction of the magnetic force acting on the straight portion of the wire and on the curved portion.

Force and Torque on a Current Loop •  The torque on a current

loop in a magnetic field is

τ = NIABsinθ

•  In vector form, where is the magnetic dipole moment of the loop.

Force and Torque on a Current Loop

µ ≡ I

A

τ = I

A ×B =µ ×B

Potential Energy of a Loop-‐‑Magnetic Field System

UB = −µ ⋅B = −µBcosθ

•  The potential energy of a system with a current-carrying loop placed in a magnetic field is given by

o  Lowest potential energy when µ and B are in the same direction. o  Highest potential energy when µ and B are oppositely directed.


A rectangular coil of dimensions 5.40 cm × 8.50 cm consists of 25 turns of wire and carries a current of 15.0 mA. A 0.350-T magnetic field is applied parallel to the plane of the coil. (a) Calculate the magnitude of the magnetic

dipole moment of the coil. (b) What is the magnitude of the torque acting on

the loop?


Consider the loop of wire shown in Figure (a). Imagine it is pivoted along side 4, which is parallel to the z axis and fastened so that side 4 remains fixed and the rest of the loop hangs vertically in the gravitational field of the Earth but can rotate around side 4 [Fig. (b)]. The mass of the loop is 50.0 g, and the sides are of lengths a = 0.200 m and b = 0.100 m.

Example 9, cont. (from Serway & JeweI)

The loop carries a current of 3.50 A and is immersed in a vertical uniform magnetic field of magnitude 0.0100 T in the positive y direction [Fig. (c)]. What angle does the plane of the loop make with the vertical?

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Physics1C Chapter 27 Slides