Download Magnetic Forces and Fields_ Induction

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
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Equations and Symbols
B
FB
q 0 v sin 
where
mv
qB
FB  ILB sin 
r
B
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 currentcarrying 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
0 I
2r
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.
F
F
N
S
v
B
I or v
2
B
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.
B
q = +1.6 x 10-19 C
v = 4.0 x 106 m/s
q
B = 0.5 T
v
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.
B
F
q

v


F  qvB  1.6 x10 19 C 4.0 x10 6 m / s 0.5T   3.2 x10 13 N
3
(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
mv 2
qvB 
r
mv 1.7 x10 27 kg 4.0 x10 6 m / s
r

 8.5 x10 2 m
19
qB
1.6 x10 C 0.5T 





(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:
B
q
v
The net force acting on the moving charge is the sum of the electric and magnetic forces,
and is called the Lorentz force:
Fnet  FE  FB
Fnet  qE  qvB
The net force in this case is zero, so the magnitude of the electric and magnetic forces are equal to each other :
qE  qvB


E  vB  4.0 x10 6 m / s 0.5T   2.0 x10 6
N
C
This expression relates the speed of the charge and the electric and magnetic fields for a
charge moving undeflected through the fields.
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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 currentcarrying 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:
F
N
S
B
B
I or v
I
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.
Top View
45
N
S
B
I
Determine the magnitude and direction of the magnetic force acting on the wire.
(sin 45 = cos 45 = 0.7)
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Solution
The magnitude of the force on the wire is found by
F  ILB sin   20 A0.10 m0.8T sin 45  1.13 N
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.
B
a
b
I
c
d
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.
Fab  ILB  2.0 A0.1m1.2T   0.24 N into the page.
Fcd  ILB  2.0 A0.1m1.2T   0.24 N 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:
  2rF  20.05 m0.24 N   0.024 Nm
6
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.
current I
I
Magnetic Field B
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
B
o I
2r
r
I
Tm
. The value 4
A
in this constant is often used for reasons of geometry. The magnetic field around a
1
current-carrying wire is proportional to , while electric field around a point charge is
r
1
proportional to 2 .
r
where o is called the permeability constant and is equal to 4 x 10-7
7
REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
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
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 4 – 6:
An electron enters a magnetic field as
shown.
e
B
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.
θ
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.
B
N
v
5.The magnitude of the force acting on
the electron is
(A) evB
(B) evB cos
(C) evBsin 
(D) vB
(E) vBsin 
S
I
6. The resulting path of the electron is a
(A) parabola
(B) straight line
(C) spiral or helix
(D) hyperbola
(E) circle
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.
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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)
P
y
r
2r
I
x
+z
(out of page)
4I
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.
P
r
v
e
I
2r
4I
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.
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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:
 I
Br  o into the page (-z)
2r
 4 I 
out of the page (+z)
B2 r  o
2 2r 
 4 I   o I  o I
Bnet  B2 r  Br  o


out of the page (+z)
2 2r  2r 2r
(b) 2 points
Let the charge on one electron be e. Then
 I 
F  evB  ev o 
 2r 
The direction of the force is up to the top of the page (+y) by the left-hand rule for a
moving negative charge.
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(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.
FB
E
FE
The magnitude of the electric field can be found by setting the magnetic force equal to
the electric force:
FB  FE
evB  eE
 I 
E  vB  v o 
 2r 
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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
12
Equations and Symbols
  vBL

where
I
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
 = 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
R
  BA cos 

  N
t
VS
N
 S
VP N P
IS NP

IP NS
P  IV
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.
13
v
L
B (out of the page)
The motional emf produced in the rod is
ε = vBL
If the circuit has a resistance R, the induced current is
 vBL
I 
R
R
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.
B
v
+q
FB
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.
14
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:
B
w
L
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:
  BA
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  relative
to the magnetic field, the flux would be
  BA cos 
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:
w
x
L
v
15
Faraday’s law of induction states that an emf  will be induced in a loop of wire through
which the flux is changing:

 
t
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

   BA 

t
t
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

 BLx 
 x 
  BL     BLv
t
 t 
The current can be found by Ohm’s law:
I

R

 BLv
R
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
16
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.
N
S
B
17
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
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.
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.
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.
18
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.
v
L
B (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
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
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
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)
a
a
h
B
19
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
  BLv  0.2 0T 0.30 m2.0 m / s   0.12V
6. A
I

R

0.12V
 0.012 A
10 
7. B
20
P  I 2 R  0.012 A 10   1.44 x10 3 Watts
2
Free Response Question Solution
(a) 3 points
Conservation of energy
U top  K bottom
mgh 
1
mv
2


v  2 gh  2 10m / s 2 1.0 m   4.5 m / s
(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.
FB  mg  1.5 kg 10 m / s 2  15 N

(d) 5 points
FB  IaB 

aB
R
FB R
a
Since FB is equal to the weight of the loop, and   BLv , we can solve for B:
B
B
FB R

a 2v
15 N 5.0   10.2 T
0.4 m2 4.5 m / s 
(e) 2 points
  Bav  10.2T 0.4 m0.45 m / s   1.8V
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