Chapter 21



MAGNETIC FORCES AND MAGNETIC FIELDS

A magnetic field is the condition of the space around a magnet in which another magnet will experience a force. Magnetic poles can be north or south, and like poles repel each other and unlike poles attract. Fundamentally, magnetism is caused by moving charges, such as a current in a wire. Thus, a moving charge or current-carrying wire produces a magnetic field, and will experience a force if placed in an external magnetic field.

QUICK REFERENCE

Important Terms

electromagnet

a magnet with a field produced by an electric current

law of poles

like poles repel each other and unlike poles attract

magnetic domain

cluster of magnetically aligned atoms

magnetic field

the space around a magnet in which another magnet or moving charge

will experience a force

mass spectrometer

a device which uses forces acting on charged particles moving a magnetic field and the resulting path of the particles to determine the relative masses of the charged particles

right-hand rules

used to find the magnetic field around a current-carrying wire or the

force acting on a wire or charge in a magnetic field

solenoid

a long coil of wire in the shape of a helix; when current is passed through a solenoid it produces a magnetic field similar to a bar magnet

Equations and Symbols

[pic]

where

B = magnetic field

FB = magnetic force

q = charge

v = speed or velocity of a charge

θ = angle between the velocity of a

moving charge and a magnetic field,

or between the length of a current-

carrying wire and a magnetic field

r = radius of path of a charge moving in

a magnetic field, or radial distance

from a current-carrying wire

m = mass

I = current

L = length of wire in a magnetic field

μ0 = permeability constant

= 4π x 10-7 (T m) / A

The Force That a Magnetic Field Exerts on a Moving Charge, The Motion of a Charged Particle in a Magnetic Field, and The Mass Spectrometer

Since a moving charge creates a magnetic field around itself, it will also feel a force when it moves through a magnetic field. The direction of the force acting on such a charge is given by the right-hand rule, with the thumb pointing in the direction of the velocity of the charge. We use our right hand for moving positive charges, and our left hand for moving negative charges.

Right-hand Rule No. 1 for force on a moving charge: Place your fingers in the direction of the magnetic field (north to south), your thumb in the direction of the velocity of a moving charge (or current in a wire), and the magnetic force on the charge (or wire) will come out of your palm.

The equation for finding the force on a charge moving through a magnetic field is

F = qvBsin (

where q is the charge in Coulombs, v is the velocity in m/s, B is the magnetic field in Teslas, and ( is the angle between the velocity and the magnetic field. If the angle is 90(, the equation becomes F = qvB.

Example 1

A proton enters a magnetic field B which is directed into the page. The proton has a charge +q and a velocity v which is directed to the right, and enters the magnetic field perpendicularly.

q = +1.6 x 10-19 C

v = 4.0 x 106 m/s

B = 0.5 T

Determine

(a) the magnitude and direction of the initial force acting on the proton

(b) the subsequent path of the proton in the magnetic field

(c) the radius of the path of the proton

(d) the magnitude and direction of an electric field that would cause the proton to continue moving in a straight line.

Solution

(a) As the proton enters the magnetic field, it will initially experience a force which is directed upward, as we see from using the right-hand rule.

[pic]

(b) The path of the proton will curve upward in a circular path, with the magnetic force becoming a centripetal force, changing the direction of the velocity to form the circular path.

(c) The radius of this circle can be found by setting the magnetic force equal to the centripetal force:

magnetic force = centripetal force

[pic]

[pic]

(d) If the charge is to follow a straight-line path through the magnetic field, we must orient the electric field to apply a force on the moving charge that is equal and opposite to the magnetic force. In this case, the electric force on the charge would need to be directed downward to counter the upward magnetic force. The electric field between the plates would be directed downward, as shown below:

The net force acting on the moving charge is the sum of the electric and magnetic forces, and is called the Lorentz force:

[pic]

This expression relates the speed of the charge and the electric and magnetic fields for a charge moving undeflected through the fields.

The Force on a Current in a Magnetic Field

Since a current-carrying wire creates a magnetic field around itself according to the first right-hand rule, every current-carrying wire is a magnet. Thus, if we place a current-carrying wire in an external magnetic field, it will experience a force. Once again, the direction of the force acting on the wire is given by the right-hand rule:

Again, you would use your left hand to find the direction of the magnetic force if you were given electron flow instead of conventional current.

The equation for finding the force on a current-carrying wire in a magnetic field is

F = ILBsin (

where I is the current in the wire, L is the length of wire which is in the magnetic field, B is the magnetic field, and ( is the angle between the length of wire and the magnetic field. If the angle is 90 (, the equation becomes simply F = ILB.

Example 2

A wire carrying a 20 A current and having a length L = 0.10 m is placed between the poles of a magnet at an angle of 45(, as shown. The magnetic field is uniform and has a value of 0.8 T.

Determine the magnitude and direction of the magnetic force acting on the wire.

(sin 45( = cos 45( = 0.7)

Solution

The magnitude of the force on the wire is found by

[pic]

The direction of the force can be found by the right-hand rule. Place your fingers in the direction of the magnetic field, and your thumb in the direction of the length (and current) which is perpendicular to the magnetic field, and we see that the force is out of the page.

Note that the length must have a component which is perpendicular to the magnetic field, or there will be no magnetic force on the wire. In other words, if the wire is placed parallel to the magnetic field, sin 0( = 0, and the force will also be zero.

Remember, use your right hand for current or moving positive charges, and your left hand for electron flow or moving negative charges.

Example 3

A wire is bent into a square loop and placed completely in a magnetic field B = 1.2 T. Each side of the loop has a length of 0.1m and the current passing through the loop is 2.0 A. The loop and magnetic field is in the plane of the page.

(a) Find the magnitude of the initial force on

each side of the wire.

(b) Determine the initial net torque acting on the

loop.

Solution

By the right-hand rule, side ab will experience a force downward into the page and side cd will experience a force upward out of the page. The current in sides bd and ac are parallel to the magnetic field, so there is no magnetic force acting on them.

[pic]into the page.

[pic] out of the page.

The result of the opposite forces on ab and cd is a torque on the loop, causing it to rotate in the magnetic field. This is the basic principle behind ammeters, voltmeters, and the electric motor. In this case, two equal and opposite forces cause the torque on the loop:

[pic]

Magnetic Fields Produced by Currents, and Ampere’s Law

A current-carrying wire creates a magnetic field around itself. Fundamentally, magnetic fields are produced by moving charges. This is why all atoms are tiny magnets, since the electrons around the nucleus of the atom are moving charges and are therefore magnetic. The magnetic field due to a current-carrying wire circulates around the wire in a direction can be found by another right-hand rule.

Right-hand Rule No. 2 for the magnetic field around a current-carrying wire:

Place your thumb in the direction of the current I, and your fingers will curl around in the direction of the magnetic field produced by that current.

In determining the direction of a magnetic field due to the flow of electrons in a wire, we would use the left hand instead of the right hand.

If the distance r from the wire is small compared to the length of the wire, we can find the magnitude of the magnetic field B by the equation

[pic]

where (o is called the permeability constant and is equal to 4( x 10-7 [pic]. The value 4( in this constant is often used for reasons of geometry. The magnetic field around a current-carrying wire is proportional to [pic], while electric field around a point charge is proportional to [pic].

REVIEW QUESTIONS

For each of the multiple choice questions below, choose the best answer.

[pic]

1. A wire carries a current, creating a magnetic field around itself as shown.

The current in the wire is

A) directed to the right.

B) directed to the left.

C) equal to the magnetic field.

D) in the same direction as the magnetic field.

E) zero.

Questions 2 – 3:

A wire carrying a current of 2 A is placed in a magnetic field

of 0.1 T as shown. The length of wire in the magnetic field

is 0.3 m.

2. The force on the wire is directed

A) into the page.

B) out of the page.

C) toward the top of the page.

D) toward the bottom of the page.

E) to the left.

3. The magnitude of the force on the wire is

A) 0.06 N

B) 2.0 N

C) 6.7 N

D) 0.15 N

E) 0.015 N

Questions 4 – 6:

An electron enters a magnetic field as shown.

[pic]

4. The electron will experience a force which is initially

(A) into the page.

B) out of the page.

C) toward the top of the page.

D) toward the bottom of the page.

E) to the left.

5.The magnitude of the force acting on the electron is

(A)[pic]

(B) [pic]

(C) [pic]

(D) [pic]

(E) [pic]

6. The resulting path of the electron is a

A) parabola

B) straight line

C) spiral or helix

D) hyperbola

E) circle

Free Response Question

Directions: Show all work in working the following question. The question is worth 10 points, and the suggested time for answering the question is about 10 minutes. The parts within a question may not have equal weight.

1. (10 points)

[pic]

Two wires cross each other at right angles. The vertical wire is carrying a current I and the horizontal wire is carrying a current 4I. Point P is a perpendicular distance r from the vertical wire, and a distance 2r from the horizontal wire.

(a) With reference to the coordinate system shown at the right, determine the magnitude and direction of the magnetic field at point P.

An electron is moving parallel to the horizontal wire with a speed v in the +x direction. Determine each of the following as the charge passes point P:

(b) the magnitude and direction of the net force acting on the electron

(c) the magnitude and direction of the electric field necessary to keep the electron moving in a straight horizontal path.

ANSWERS AND EXPLANATIONS TO REVIEW QUESTIONS

Multiple Choice

1. A

By right-hand rule no. 2, if the fingers curl in the direction of the magnetic field around the wire (up and over the wire toward you), the current must be to the right.

2. B

According to right-hand rule no. 1, the fingers point to the right in the direction of the magnetic field, the thumb points toward the bottom of the page in the direction of the current, and the force comes out of the palm and out of the page.

3. A

F = ILB = (2 A)(0.1 T)(0.3 m) = 0.06 N

4. B

We use the left-hand-rule, since the electron is a negative charge, placing our fingers toward the bottom of the page, and the thumb to the right, in the direction of the component of the velocity which crosses the magnetic field lines. The force, then, comes out of the palm and out of the page.

5. C

The magnitude of the force is equal to the product of the charge, speed, and magnetic field, and the sine of the angle between the velocity and the magnetic field, which is θ.

6. C

The electron will orbit a magnetic field line, but also continue to move down toward the bottom of the page, therefore spiraling downward.

Free Response Question Solution

(a) 4 points

The net magnetic field at point P is due to the magnetic fields produced by both wires:

[pic] into the page (-z)

[pic] out of the page (+z)

[pic] out of the page (+z)

(b) 2 points

Let the charge on one electron be e. Then

[pic]

The direction of the force is up to the top of the page (+y) by the left-hand rule for a moving negative charge.

(c) 4 points

In order to keep the electron moving in a straight path to the right, we would need to apply a downward (-y) electric force to the electron which is equal and opposite to the upward (+y) magnetic force.

Since electric field lines are drawn in the direction a

positive charge would experience a force, they would

be drawn in the opposite direction an electron would

experience a force. Thus, the electric field would need

to be applied in the +y direction to keep the electron

moving in a straight horizontal line.

The magnitude of the electric field can be found by setting the magnetic force equal to the electric force:

[pic]

ELECTROMAGNETIC INDUCTION

Since a current produces a magnetic field, we may assume that a magnetic field can produce a current. Electromagnetic induction is the process by which an emf (or voltage) is produced in a wire by a changing magnetic flux. Magnetic flux is the product of the magnetic field and the area through which the magnetic field lines pass. Electromagnetic induction is the principle behind the electric generator and the transformer. The direction of the induced emf or current is governed by Lenz’s law.

QUICK REFERENCE

Important Terms

alternating current

electric current that rapidly reverses its direction

electric generator

a device that uses electromagnetic induction to convert mechanical

energy into electrical energy

electromagnetic induction

inducing a voltage in a conductor by changing the magnetic

field around the conductor

induced current

the current produced by electromagnetic induction

induced emf

the voltage produced by electromagnetic induction

Faraday’s law of induction

law which states that a voltage can be induced in a

conductor by changing the magnetic field around the conductor

Lenz’s law

the induced emf or current in a wire produces a magnetic flux which opposes

the change in flux that produced it by electromagnetic induction

magnetic flux

the product of the magnetic field and the area through which the magnetic

field lines pass.

motional emf

emf or voltage induced in a wire due to relative motion between the wire and a magnetic field

Equations and Symbols

[pic]

where

ε = emf (voltage) induced by

electromagnetic induction

v = relative speed between a conductor

and a magnetic field

B = magnetic field

L = length of a conductor in a magnetic

field

I = current

R = resistance

Φ = magnetic flux

A = area through which the flux is

passing

[pic] = angle between the direction of the

magnetic field and the area through

which it passes

N = number of loops in a coil of wire

VS = voltage in the secondary coil of a

transformer

VP = voltage in the primary coil of a

transformer

NS = number of loops in the secondary

coil of wire in a transformer

NP = number of loops in the primary coil

of wire in a transformer

Motional Emf

In the 1830’s Faraday and Henry independently discovered that an electric current could be produced by moving a magnet through a coil of wire, or, equivalently, by moving a wire through a magnetic field. Generating a current this way is called electromagnetic induction.

If we move a rod perpendicular to a magnetic field, there is a magnetic force on the charges in the rod, sending the positive charges to one end of the wire and the negative charges to the other, as shown in your textbook. This polarization of charge creates a potential difference or emf ε between the ends of the rod. If the rod closes a loop, the induced ε produces a current I around the closed loop.

[pic]

The motional emf produced in the rod is

ε = vBL

If the circuit has a resistance R, the induced current is

[pic]

Example 1

In the diagram above, the magnetic field is directed out of the page, and the rod is moving to the right. Find the direction of the induced current in the rod.

Solution

Consider a small positive charge inside the rod. As the charge moves to the right with the rod, we can find the direction of the magnetic force on the charge by using right-hand rule no. 2.

[pic]

Since the force on a positive charge in the rod is downward toward the bottom of the page, the flow of positive charge and thus the induced current is toward the bottom of the page.

Magnetic Flux and Faraday’s Law of Electromagnetic Induction, and

Lenz’s Law

Consider a rectangular loop of wire of height L and width x which sits in a region of magnetic field of length 3x. The magnetic field is directed into the page, as shown below:

The loop is in the plane of the page so that its area is perpendicular to the magnetic field lines. The magnetic flux ( through the loop is the scalar product of the magnetic field B and the area A through which it fluxes:

[pic]

As the loop sits a rest in the magnetic field, the magnetic flux through it is constant, and is equal to BA = BLx. If the loop were tilted forward (toward you) at an angle [pic] relative to the magnetic field, the flux would be

[pic]

But if the loop is moved through the magnetic field with a constant speed v, the flux through the loop does not remain constant, but changes with time:

Faraday’s law of induction states that an emf ( will be induced in a loop of wire through which the flux is changing:

[pic]

Note that the induced emf in the wire is not caused by the magnetic flux or field, but by the change in flux through the loop.

It may occur to you that we could use Faraday’s law of induction to create an emf or current, and the induced current would produce a new magnetic field and flux which could produce more current, and so on. But Lenz’s law tells us that this is not possible.

Lenz’s law states that the induced emf or current in a wire produces a change in flux which opposes the change in flux that produced it.

The negative sign on the equation for the induced emf above tells us that the new change in flux works against the old change in flux. This is essentially a statement of conservation of energy.

The the emf induced in the loop as the loop is passing through the magnetic field is

[pic]

where A is the area of magnetic field which is enclosed by the loop of wire. In the position of the loop shown, this area would be Lx. So, the equation for ( becomes

[pic]

The current can be found by Ohm’s law:

[pic]

Faraday’s law of induction is illustrated in the excellent examples in your textbook, and in the review questions and free response problem that follow.

The Electric Generator

The principle of electromagnetic induction is the basis for a generator. A generator converts mechanical energy into electrical energy. Place a loop of wire on an axle in a magnetic field. As the loop is rotated, the wire crosses magnetic field lines and generates a current in the loop. That current can be used to light a light bulb, or power a city. All of our electrical power is generated in a similar way.

CHAPTER 22 REVIEW QUESTIONS

For each of the multiple choice questions below, choose the best answer.

1. All of the following can induce and emf in a coil of wire EXCEPT

(A) moving a magnet through the coil

(B) placing a stationary coil of wire in an

increasing magnetic field

(C) placing a stationary coil of wire in a

stationary magnetic field.

(D) placing a stationary coil of wire in

an decreasing magnetic field

(E) moving both the coil of wire and a

magnet away from each other at the

same speed

2. A magnet moves into a coil of wire inducing a current in the wire. If the magnet pulled back out of the coil in the opposite direction as it went into the coil, which of the following will occur?

(A) There will be a current produced in

the coil in the same direction as

before.

(B) There will be a current produced in

the coil in the opposite direction as

before.

(C) There will be no current produced in

the coil.

(D) The current produced must be

stronger than before.

(E) The current produced must be weaker than before.

3. A generator

A) converts mechanical energy into electrical energy.

B) converts electrical energy into mechanical energy.

C) converts heat energy into mechanical energy.

D) converts electrical energy into heat energy.

E) converts nuclear energy into heat energy.

4. Which of the following is implied by Lenz’s law?

(A) The induced emf always opposes the

emf that produced it.

(B) The induced current always opposes

the current that produced it.

(C) The induced emf always opposes the

change in flux that produced it.

(D) The induced emf always opposes the

flux that produced it.

(E) The induced emf always opposes the

magnetic field that produced it.

Questions 5 – 7: A conducting rod of length 0.30 m and resistance 10.0 Ω moves with a speed of 2.0 m/s through a magnetic field of 0.20 T which is directed out of the page.

5. The emf induced in the rod is

(A) 0.12V

(B) 0.40V

(C) 0.60 V

(D) 0.72 V

(E) 4.0 V

6. The current produced in the rod is

(A) 0.012 A

(B) 0.040 A

(C) 0.060 A

(D) 0.072 A

(E) 0.40 A

7. The power dissipated in the rod is

(A) 4.00 x 10-3 Watts

(B) 1.44 x 10-3 Watts

(C) 6.14 x 10-3 Watts

(D) 3.14 x 10-3 Watts

(E) 2.33 x 10-3 Watts

Free Response Question

Directions: Show all work in working the following question. The question is worth 15 points, and the suggested time for answering the question is about 15 minutes. The parts within a question may not have equal weight.

1. (15 points)

[pic]

A square loop of sides a = 0.4 m, mass m = 1.5 kg, and resistance 5.0 Ω falls from rest from a height h = 1.0 m toward a uniform magnetic field B which is directed into the page as shown.

(a) Determine the speed of the loop just before it enters the magnetic field.

As the loop enters the magnetic field, an emf ε and a current I is induced in the loop.

(b) Is the direction of the induced current in the loop clockwise or counterclockwise?

Briefly explain how you arrived at your answer.

When the loop enters the magnetic field, it falls through with a constant velocity.

(c) Calculate the magnetic force necessary to keep the loop falling at a constant velocity.

(d) What is the magnitude of the magnetic field B necessary to keep the loop falling at a

constant velocity?

(e) Calculate the induced emf in the loop as it enters and exits the magnetic field.

ANSWERS AND EXPLANATIONS TO REVIEW QUESTIONS

Multiple Choice

1. E

There must be relative motion between the coil and magnet for an emf to be induced.

2. B

Pushing the north end of a magnet into a coil will produce a current in the opposite direction than if the north end were pulled out of the coil. Opposite magnet velocities will create opposite currents.

3. A

Mechanical energy is put into a generator, and electrical energy is produced in the form of current.

4. C

Induced emf opposes the change in flux that produced it, not necessarily the initial flux, magnetic field, current, or emf.

5. A

[pic]

6. A

[pic]

7. B

[pic]

Free Response Question Solution

(a) 3 points

Conservation of energy

[pic]

(b) 2 points

As the positive charges in the leading edge of the loop move downward through the magnetic field, they experience a force to the right. Thus, the current moves to the right in the bottom of the loop, and around the loop counterclockwise.

(c) 3 points

The magnetic force upward just balances the weight of the loop downward.

[pic]

(d) 5 points

[pic]

[pic]

Since FB is equal to the weight of the loop, and [pic], we can solve for B:

[pic]

(e) 2 points

[pic]

-----------------------

I or v

F

B

B

v

S

N

F

v

B

q

q

B

v

F

q

B

v

B

I

S

N

I or v

F

B

Top View

45(

S

N

B

I

B

a

b

c

d

I

Magnetic Field B

current I

I

I

r

[pic]

I

4I

2r

r

P

x

y

+z

(out of page)

P

I

4I

2r

r

e

v

FB

FE

E

B

I

S

N

¸

e

v

B

v

B (out of the page)

L

+q

d B

current I

I

I

r

[pic]

I

4I

2r

r

P

x

y

+z

(out of page)

P

I

4I

2r

r

e

v

FB

FE

E

B

I

S

N

θ

e

v

B

v

B (out of the page)

L

+q

v

B

FB

w

L

B

w

L

x

v

N

S

B

v

B (out of the page)

L

a

a

B

h

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