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Honors/AP Physics: An Electronic TextbookChapter 13
ElectromagnetismMaking Electricity Work for You
© 2009 Stan Eisenstein
Lesson 13.4
Electromagnetic Induction
Return to Main Table of Contents
Lesson 13.4 Table of Contents
We have learned about two aspects of magnetism.
Each aspect has its own equations and rules for
determining direction.
Introduction
v
B F+V
FI
Aspect 1: CurrentCreates B Field
Aspect 2: B Field Exerts Force on Current and Moving Charge
Author Schorschi2; Public Domain Author Stan Eisenstein
Introduction
Aspect 1 is that current creates a B field. The B
field lines near a long straight wire form circles. The
direction of the those circles can be determined
using the Right Hand Rule:
If you point the thumb of your right hand in the direction of the current, then your fingers will curl in the direction of the magnetic field lines.
Author Schorschi2; Public Domain
Introduction
The size of the B field produced by a long straight
wire is given by the equation:
B = m0I/2pR or B = kI/R
Where ‘I’ represents the size of the current, ‘R’
represents the distance from the wire, and m0 represents a constant equal to 4p x 10-7 Tesla·meters/Ampere.
k = m0/2p = 2.0 x 10-7 T·m/A
Introduction
A coil of wire, called a solenoid, creates a strong
uniform field inside the coil and a weak, near-zero,
field outside the coil.
+-
x x x x xx
● ● ● ● ● ●
B field
Author Zureks Public Domain
Introduction
Aspect 2: Currents and charges moving through
B fields experience forces. The direction of the force
can be determined using the Right Hand Rule:
Thumb = v, I Fingers = B
Palm = F on I, + Back = F on -
+
VF
I
Author Stan Eisenstein
v
BF
Introduction
Aspect 2: The size of the force is given by these
equations:
F = qvBsinq F = ILBsinq
+
VF
I
Author Stan Eisenstein
v
BF
Introduction
Aspect 2: Charge particles moving through a B field
move in a circular or spiral path (or a straight line, if
v is parallel to B)X X X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X+ V
+
V
+
VF
F
F
R
+V┴
V║
+
Introduction
Aspect 2: The forces on a current loop in a magnetic
field causes rotation. One important application is an
electric motor.
N S
A
C
View from the Side
x
N SA
B
C
D
View from Above
x●
● ●● ●● ●● ●● ●A B
0.20 m
2.0 A
0.10 m
F
Bwire B
Iwire
Introduction
It is important to be clear about which aspect of
magnetism is being asked about in problems. We
have already seen one problem in which both Aspects 1 and 2 are included. One current
created a B field (Aspect 1) and a second current in that
field experienced a force (Aspect 2).
I
B
F
Author Schorschi2; Public DomainAuthor Stan Eisenstein
Introduction
In this lesson we introduce a 3rd Aspect of magnetism: the generation of a current using magnetism, a process called “Electromagnetic Induction”.
X X X X X X X
X X X X X X X
X X X X X X X
X X X X X X X
Introduction
The process of Electromagnetic Induction is at the
heart of such applications as 1) electric generators,
2) transformers, and 3) microphones.
Author - http://commons.wikimedia.org/wiki/File:Hoover_Dam_turbines.JPG; Public Domain Author Stan EisensteinAuthor Stan Eisenstein
Consider a wire that is pushed across a magnetic
field. Consider that there are positive charges in the
wire crossing the field.
Question 1: Use the Right Hand Rule to determine
how those charges are affected.
Electromagnetic Induction
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+
+ v
+
Electromagnetic Induction
Answer 1: The charges are forced downward.
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vB
F
F+
+ v
+ Author Stan Eisenstein
Electromagnetic Induction
The result of these charges being forced downward
is that the bottom of the wire is charged positive and
the top is charged negative.
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F
-
v
+
Electromagnetic Induction
This charge separation creates an electric potential
difference (i.e. a voltage). We call this potential
difference an induced Electromotive Force (EMF).
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F
-
v
+
E
Electromagnetic Induction
If the ends of the wire are connected by a conductor,
then a complete circuit is created and a current is
generated.
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F
-
v
+
EI
Electromagnetic Induction
It doesn’t matter whether the wire loop is moving
across the field, or whether the field is moving across the wire loop, the result is the same –
an induced EMF generating a current.
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● ● ● ● ● ● ●
F
-
v
+
I E
Electromagnetic Induction
Once the loop is entirely inside the B field, each of
the wires moving across the field experiences forces
on the charges within them. An equal EMF of same
polarity is generated in each wire. The result is that
no current is generated, though one side of the loop
is positive and the other negative.
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vF
-
+
EF
-
+
-- -
+ + +
I = 0
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● ● ● ●
In the 1830’s Michael Faraday derived an equation
that described the induced EMF in a wire loop under
various conditions: Moving a loop or coil through a
B field; changing the strength of a B field; rotating a
coil within a B field. Click to animate
Faraday’s Law
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●
x
A
C
Faraday’s Law states that the induced EMF is equal
to the number of loops in a coil times the rate at
which the “Magnetic Flux” changes.
E = N(DFB/Dt)
Where E is the induced EMF, N is the number of
loops in a coil, FB is the magnetic flux (defined in the
next slides), t is time.
Faraday’s Law
Before we can use Faraday’s Law, we must first
define “Magnetic Flux”. Magnetic Flux can be considered to be a measure of the amount of B field that passes through an area.
Faraday’s Law
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●
x
A B C D
In case A, a wire rectangle lies in the plane of the
screen/page. To get a sense of the meaning of Magnetic Flux, we can count the number of B
field lines passing through the loop. For case A
there are 12 lines passing through the loop.
Faraday’s Law
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●
x
A B C D
One way to change the Magnetic Flux is to change
the area of the loop. In case B, a loop with a smaller
area is in the same B field. This time there are only 6
B field lines passing through the loop. The flux is
reduced compared to case A.
Faraday’s Law
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A● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●
B
Flux a Area
Another way to change the Magnetic Flux is to change the strength of the B field. In case C, a
loop with the same area is in a weaker B field (as depicted by the field line density). This time
there are only 4 B field lines passing through the
loop. The flux is reduced compared to case A.
Faraday’s Law
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A C
Flux a B Field
A third way to change the Magnetic Flux is to change orientation between the loop and the
B field. In case D, a loop with the same area is in the
same B field. The loop has been rotated, with the
top of the loop moved toward you and the bottom
moved away. There are 6 B
field lines passing through the loop. The flux is reduced compared to case A.
Faraday’s Law
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A● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●
●
x
D
Flux a Orientation
If we view the loop and field from the side, we see
that the flux is a maximum when the plane of the
loop is perpendicular to the field (A). It is a minimum
(i.e. zero) when the plane of the loop is parallel to
the field (E).
Faraday’s Law
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A● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●
●
x
DA D E
In order to quantify the orientation between the loop
and the field, we define a “direction” for an Area
“vector” to be perpendicular to the plane of the area.
In case A, the Area Vector points right (or left). In
case E, the Area Vector points down (or up).
Faraday’s Law
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A● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●
●
x
DA D E
Area Vector
Magnetic Flux is defined by the equation:
FB = BAcosq
where FB represents magnetic flux, B represents the
strength of the field, A represents the area of the
loop, and q represents the angle between the Area
and B field vectors.
Faraday’s Law
FB a A FB a OrientationFB a B
FB = BAcosq
We see that this equation corresponds to the proportionality relationships we explored
when counting B field lines passing through the
loop. As B increases, FB increases.
Faraday’s Law
FB a B
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A C
FB = BAcosq
As A increases, FB increases.
Faraday’s Law
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A● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●
B
FB a A
FB = BAcosq
As q increases from 0° to 90°, FB decreases from a
maximum to zero.
FB a Orientation
Faraday’s Law
A D E
0°
90°
FB = BAcosq
Question 2: What is the unit for magnetic flux?
Faraday’s Law
FB a A FB a OrientationFB a B●
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● ●
FB = BAcosq
Answer 2: Tesla·meter2. This unit is also called a
Weber. In fact, sometimes B field strength is given
the unit Weber/meter2, which is the same unit as a
Tesla.
Faraday’s Law
FB a A FB a OrientationFB a B●
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● ●
E = N(DFB/Dt) FB = BAcosq
Faraday’s Law states that induced EMF is directly
proportional to the number of loops in a coil and to
the rate at which the magnetic flux changes. The
faster the flux changes, the greater the EMF.
Faraday’s Law
Electromagnetic Induction (1:27) Turn Sound On; Click on Video Frame to Start
Demonstration
E = N(DFB/Dt) FB = BAcosq
Question 3: Based on Faraday’s Law, what is the unit for EMF?
Faraday’s Law
E = N(DFB/Dt) FB = BAcosq
Answer 3: Tesla·meter2/second. This unit is equivalent to a volt.
B = F/qv Tesla = Newton·second/coulomb·meter
Tesla·meter2/second =
(Newton·second/coulomb·meter)(meter2/second) =
Newton·meter/coulomb = Joule/coulomb = volt
Faraday’s Law
E = N(DFB/Dt) FB = BAcosq
Question 4: For a given coil of wire, what are the
three different ways that you can generate an EMF?
Faraday’s Law
Author - http://commons.wikimedia.org/wiki/File:Hoover_Dam_turbines.JPG; Public Domain
E = N(DFB/Dt) FB = BAcosq
Answer 4: An EMF is generated by changing FB.
This is done by a) changing B, b) changing A or
c) changing q. Let’s explore each of these scenarios.
Faraday’s Law
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●
x
A B C D
E = N(DFB/Dt) FB = BAcosq
Question 5: A rectangular coil of wire with 50 loops
has a width of 10 cm, a length of 20 cm, and a resistance of 50 W. The coil is placed in a B
field that rises from 2 T to 6 T in 5 seconds. The Area vector of the coil points parallel to the field. Click to animate.
(Question continued on next slide)
Using Faraday’s Law
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10 cm
20
cm2 T6 T
E = N(DFB/Dt) FB = BAcosq
Question 5:
a) What are the initial and final values of the magnetic flux?
b) What is the size of the induced EMF?c) What is the size of the
induced current?
Using Faraday’s Law
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10 cm
20
cm
2 6 T
5 s
50 loops
50 W
E = N(DFB/Dt) FB = BAcosq
Answer 5: a) 0.04 T·m2 and 0.12 T·m2; b) 0.8 v; c) 0.016 A
FBi = BiAcosq = (2 T)(0.20 m)(0.10 m)(cos 0°) = 0.04 T·m2
FBf = (6 T)(0.20 m)(0.10 m)(cos 0°) = 0.12 T·m2
E = N(DFB/Dt) = (50)(0.12 – 0.04 T·m2)/(5 s) = 0.8 v
I = E/R = (0.8 v)/(50 W) = 0.016 A
Using Faraday’s Law
E = N(DFB/Dt) FB = BAcosq
Question 6: A single wire loop, 20 cm long and
10 cm wide, is pushed into a 5 T B field at a rate of
2 cm/s. What is the size of the induced EMF in the
loop a) before it fully enters the field? b) after it is
fully in the field but still moving? Click to animate.
Using Faraday’s Law
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5 T
10 cm
20
cm 2 cm/s
E = N(DFB/Dt) FB = BAcosq
Answer 6: a) 0.02 v; b) 0.0 v
There are two methods for solving part a)
Method 1: First, determine the flux before entering
the field (0 T·m2) and after fully entering the field:
FBf = (5 T)(0.02 m2)(cos 0°)
FBf = 0.10 T·m2
Using Faraday’s Law
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● ● ● ●
5 T
10 cm
20
cm 2 cm/s
E = N(DFB/Dt) FB = BAcosq
Answer 6: a) 0.02 v; b) 0.0 v
Method 1: FBf = 0.10 T·m2 FBi = 0.00 T·m2
Next – determine the time it takes the loop to fully
enter the field.
v = d/t t = d/v
t = (10 cm)/(2 cm/s) = 5 s
Using Faraday’s Law
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5 T
10 cm
20
cm 2 cm/s
E = N(DFB/Dt) FB = BAcosq
Answer 6: a) 0.02 v; b) 0.0 v
Method 1: FBf = 0.10 T·m2 FBi = 0.00 T·m2 t = 5 s
Finally, use Faraday’s Law to determine the EMF.
E = N(DFB/Dt)
E = (1)(0.10 – 0.00 T·m2)/(5 s)
E = 0.02 v
Using Faraday’s Law
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● ● ● ●
● ● ● ●
5 T
10 cm
20
cm 2 cm/s
E = N(DFB/Dt) FB = BAcosq
Answer 6: a) 0.02 v; b) 0.0 v
For Method 2 we will derive a generic equation for
the EMF of a wire moving through a B field. ‘L’ represents the length of the wire crossing the field and ‘W’ represents the width of the loop already in the field.
Using Faraday’s Law
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● ● ● ●
● ● ● ●
B
W
L v
E = N(DFB/Dt) FB = BAcosq
Answer 6: a) 0.02 v; b) 0.0 v
Method 2: The first step is to determine an expression for FB at a given moment.
FB = BAcosq = BLw
Using Faraday’s Law
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● ● ● ●
● ● ● ●
B
W
L v
E = N(DFB/Dt) FB = BAcosq
Answer 6: a) 0.02 v; b) 0.0 v
Method 2: FB = BLw The next step is use Faraday’s Law to
determine the EMF.
E = N(DFB/Dt) = (1)( D (BLw) /Dt)
B and L are constant.
E = BL( D w/Dt) = BLv
Using Faraday’s Law
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● ● ● ●
● ● ● ●
B
W
L v
E = N(DFB/Dt) FB = BAcosq
Answer 6: a) 0.02 v; b) 0.0 v
Method 2: E = BLv
This equation holds for wires moving across B fields.
E = (5 T)(0.20 m)(0.02 m/s)
E = 0.02 v
Using Faraday’s Law
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5 T
0.2
0 m 0.02 m/s
E = N(DFB/Dt) FB = BAcosq
Answer 6: a) 0.02 v; b) 0.0 v
b) Once the loop is entirely within the field, FB is constant; it is no longer changing. This makes the induced EMF equal to zero.
E = (1)(0.10 – 0.10 T·m2)/(5 s) = 0 volts
Using Faraday’s Law
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● ● ● ●
E = N(DFB/Dt) FB = BAcosq
Answer 6: a) 0.02 v; b) 0.0 v
b) In actuality, each wire crossing the field has an EMF induced equal to BLv. But since these EMFs oppose each other, the net EMF is zero. Charge will separate to opposite ends of the loop, but no current will flow.
Using Faraday’s Law
● ● ● ●
● ● ● ●
● ● ● ●
● ● ● ●
B
L v
-
+
--
+ +
We have already learned one method of determining
the direction of induced current – the Right Hand
Rule. As a wire crosses a B field, we consider the
force on the + charges in the wire. This force is the
direction of the current in the wire.
Direction of Induced Current
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
F v+
EI ++
vB
FAuthor Stan Eisenstein
Question 7: As the wire loop enters the field shown
here, the current is induced in a clockwise direction.
Use the Right Hand Rule to determine the direction
of the current as the loop leaves the field.
Direction of Induced Current
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
F v+
EI ++
v
Answer 7: Counterclockwise. The force on the left-
hand wire is down (just like the right-hand wire as
the loop enters the field). This downward force
causes a CCW current.
Direction of Induced Current
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
FI
vvB
FAuthor Stan Eisenstein
Unfortunately, the Right Hand Rule is not easy to
apply in all induction situations. For instance, if the
wire loop does not move, but the B field changes
size, then there is no velocity and the Right Hand
Rule becomes difficult or impossible to use.
Direction of Induced Current
● ● ● ●
● ● ● ●
● ● ● ●
● ● ● ●
10 cm
20
cm
2 6 T
5 s
50 loops
50 W
?B
?Author Stan Eisenstein
A second rule that may be used to determine direction of induced current is the “Left Hand
Rule”. This rule was made up for this text and is not generally recognized. However, this rule will accurately determine the direction of induced
current for all situations.
Direction of Induced Current
● ● ● ●
● ● ● ●
● ● ● ●
● ● ● ●
10 cm
20
cm
2 6 T
5 s
50 loops
50 W
Author Schorschi2 at de.wikipedia; Public Domain; Adapted
Left Hand Rule:
If Flux is increasing, point Thumb in direction of B
If Flux is decreasing, point Thumb opposite B Fingers curl in the direction of the current.
Direction of Induced Current
● ● ● ●
● ● ● ●
● ● ● ●
● ● ● ●
10 cm
20
cm
2 6 T
5 s
50 loops
50 W
Author Schorschi2 at de.wikipedia; Public Domain; Adapted
If FB ↑, Thumb = B; If FB ↓, Thumb opp. B; Fingers = I.
Question 8a: Use the Left Hand Rule to determine
the direction of the current in the wire loop shown
below as it enters and leaves the magnetic field.
Direction of Induced Current
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
v v
If FB ↑, Thumb = B; If FB ↓, Thumb opp. B; Fingers = I.
Answer 8a: Clockwise; Counterclockwise
Entering: As the area ↑, FB ↑; Point Thumb in direction of B – toward you. Fingers curl CW.
Direction of Induced Current
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
vI
Author Stan Eisenstein
B
I
If FB ↑, Thumb = B; If FB ↓, Thumb opp. B; Fingers = I.
Answer 8a: Leaving: As the area ↓, FB ↓; Point Thumb opp direction of B – away. Fingers curl
CCW.
Direction of Induced Current
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
vI
B
I
Author Stan Eisenstein
If FB ↑, Thumb = B; If FB ↓, Thumb opp. B; Fingers = I.
Answer 8a: These directions match the directions
determined by the Right Hand Rule.
Direction of Induced Current
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
vI
vI
If FB ↑, Thumb = B; If FB ↓, Thumb opp. B; Fingers = I.
Question 8b: For the loop shown below, the B field
changes from 5 T away to 5 T toward you. Determine the direction of the current while
the field away from you decreases and then again
while the field toward you increases.
Direction of Induced Current
X X X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X
If FB ↑, Thumb = B; If FB ↓, Thumb opp. B; Fingers = I.
Question 8b: Clockwise for both. While the field and
flux are decreasing away from you, the thumb points
opp. B (toward you). While the field and flux are increasing toward you, the thumb points with the
field, again toward you.
Direction of Induced Current
X X X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X
Opp.B B
Author Stan Eisenstein
I
Author Stan Eisenstein
I
If FB ↑, Thumb = B; If FB ↓, Thumb opp. B; Fingers = I.
Question 8c: A coil of wire is rotating in a clockwise
direction in a field pointing to the right. Determine
the direction of the current in wires A and C for
positions 1 and 2 of the coil in its rotation.
Direction of Induced Current
A A
C C
1 2
If FB ↑, Thumb = B; If FB ↓, Thumb opp. B; Fingers = I.
Answer 8c: The currents are shown in the diagram.
At position 1, the flux is increasing. The Thumb
points Right and the fingers curl away at A and
toward you at C. At position 2, the flux is decreasing.
The Thumb points left and the fingers curl toward
you at A and away at C.
Direction of Induced Current
A A
C C
1 2
x
x ●
●
1 2A A
CC
Author Schorschi2 at de.wikipedia; Public Domain; Adapted
If FB ↑, Thumb = B; If FB ↓, Thumb opp. B; Fingers = I.
Question 8d: A wire with an increasing current lies
next to a rectangular loop of wire. What is the direction of the induced current in the loop?
Direction of Induced Current
If FB ↑, Thumb = B; If FB ↓, Thumb opp. B; Fingers = I.
Answer 8d: CW.
There are two parts to this problem: Aspect 1 – a
current creates a B field; Aspect 3 – a changing flux induces a current.
Direction of Induced Current
Author Schorschi2 at de.wikipedia; Public Domain; Adapted Author Stan Eisenstein
I
If FB ↑, Thumb = B; If FB ↓, Thumb opp. B; Fingers = I.
Answer 8d: CW.
Using the Right Hand Rule, we find that the wire with the current creates a B field toward you at the position of the loop.
Direction of Induced Current
●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●Author Schorschi2 at de.wikipedia; Public Domain; Adapted
If FB ↑, Thumb = B; If FB ↓, Thumb opp. B; Fingers = I.
Answer 8d: CW.
Since the current is increasing, the strength of the
field and the flux are increasing. The Left Hand Rule shows that the induced current must be CW.
Direction of Induced Current
BI
●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●Author Stan Eisenstein
The more traditional method of determining the
direction of induced current is called “Lenz’s Law”.
Lenz’s Law states that a current is induced in a
direction that resists the change in flux.
Lenz’s Law
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
vI
vI
Let’s explore Lenz’s Law in some of the examples
we have already discussed.
For a loop being pushed into a B field, the FB is clearly increasing. Lenz’s Law states that the induced current should resist this increase in
FB. How might this occur?
Lenz’s Law
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
vI A
We know from the Right and Left Hand Rules that a
CW current is induced in the loop. This is according
to Aspect 3 – Electromagnetic Induction.
Now consider Wire A. It conducts a current across a
B field. Consider Aspect 2 of Magnetism – a B field
exerts a force on a Current. The B field will exert a
force on this induced current.
Lenz’s Law
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
vI A
Question 9a: Use the Right Hand Rule to determine
the direction of the force of the B field on the current
in Wire A.
Lenz’s Law
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
vI A
Answer 9a: Left.
The fingers point toward you, in the direction of the
B field. The Thumb points down, in the direction of
the current. The palm points Left, in the direction of
the Force on the current.
Lenz’s Law
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
vI FB
BF
IAuthor Stan Eisenstein
Answer 9a: Left.
Notice that as the loop tries to enter the field and
increase the flux, the field tries to push the loop back
out, resisting the increase.
Lenz’s Law
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
vI FB
BF
IAuthor Stan Eisenstein
Question 9b: Now consider as the loop leaves the
field. We have already determined that the induced
current flows CCW. Use the Right Hand Rule to determine the direction of the force that the B
field exerts on the current in wire C.
Lenz’s Law
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
v
I
C
Answer 9b: Left. The force points in the same direction that it did when the loop was
entering the field. Now the flux is reducing as the loop
leaves the field. The B field exerts a force that resists the reduction of flux – it tries to pull the loop back
into the field.
Lenz’s Law
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
v
I
CF
BF
IAuthor Stan Eisenstein
Lenz’s Law relates to the conservation of energy.
The force from the field must oppose the velocity. If
the force were in the same direction as the velocity,
the speed would keep getting faster, creating Kinetic
Energy out of nothing.
Lenz’s Law
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
vI
vIFF
Consider now a loop in a B field that points away
that is decreasing in strength. We have already
determined that a clockwise current is induced.
Since the flux is reducing, the induced current must
somehow try it resist the change. The new current
creates its own B field within the loop (Aspect 1).
Lenz’s Law
X X X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X
B decreasing
Question 9c: Use the Right Hand Rule to determine
the direction of the B field created by the induced
current.
Lenz’s Law
X X X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X
B decreasing
Answer 9c: The induced current generates a field
away from you inside the loop.
As the decreasing B field reduces the flux, the induced current generates its own field to
reinforce the original field, resisting the decrease in flux.
Lenz’s Law
X X X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X
B decreasing
x x x x
x x xx
Author Schorschi2 at de.wikipedia; Public Domain
Question 10: A conducting bar moves frictionlessly
on conducting rails in a 5 T field point away. A 10 W
resistor connects the rails at the left end (the rest of
the circuit has minimal resistance). The bar is pushed so that it moves at a constant speed
of 5 m/s. (continued on next slide)
Practice Problem
x x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x
10 W
20 cm
5 m/s5 T
1 m 10 m
Question 10: The rails are 20 cm apart. The bar
starts 1 meter from the left end of the rails and
moves 10 meters. a) What is the direction of the induced current
in the moving bar?(continued on next slide)
Practice Problem
x x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x
10 W
20 cm
5 m/s5 T
1 m 10 m
Question 10: b) What is the size of the EMF induced in the
bar?c) What is the size of the current?d) What is the force needed to move the bar
at a constant speed?(continued on next slide)
Practice Problem
x x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x
10 W
20 cm
5 m/s5 T
1 m 10 m
Question 10: e) How much work is done moving the bar 10
m?f) How much thermal energy is generated in
the resistor in the time it takes to move the bar
10 m?
Practice Problem
x x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x
10 W
20 cm
5 m/s5 T
1 m 10 m
Answer 10: a) The current in the bar is up. LH Rule: Thumb = Away (Flux increases)
Fingers = I = CCW
Practice Problem
x x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x
10 W
20 cm
5 m/s5 T
1 m 10 m
Answer 10: b) 5 volts. Method 1: E = BLv = (5 T)(0.20 m)(5 m/s) = 5
v
Practice Problem
x x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x
10 W
20 cm
5 m/s5 T
1 m 10 m
b) 5 volts. Method 2: FBi = BA = (5 T)(0.20 m)(1 m) = 1
T·m2
FBf = BA = (5 T)(0.20 m)(11 m) = 11 T·m2
t = d/v = (10 m)/(5 m/s) = 2 sE = /DF Dt = (11 – 1 T·m2)/2 s = 5 v
Practice Problem
x x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x
10 W
20 cm
5 m/s5 T
1 m 10 m
Answer 10: c) 0.5 amperes.
I = E/R = (5 v)/(10 W) = 0.5 A
Practice Problem
x x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x
10 W
20 cm
5 m/s5 T
1 m 10 m
Answer 10: d) 0.5 newtons.
F = ILB = (0.5 A)(0.20 m)(5 T) = 0.5 n
Practice Problem
x x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x
10 W
20 cm
5 m/s5 T
1 m 10 m
Answer 10: e) 5 joules.
W = Fd = (0.5 n)(10 m) = 5 joules
Practice Problem
x x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x
10 W
20 cm
5 m/s5 T
1 m 10 m
Answer 10: f) 5 joules.
E = Pt = IVt = (0.5 A)(5 v)(2 s) = 5 joules
Practice Problem
x x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x
10 W
20 cm
5 m/s5 T
1 m 10 m
W = Fd = 5 Joules E = IVt = 5 joules
Note that the work done in moving the bar is converted to electric energy and ultimately
thermal energy in the resistor. Energy is conserved.
Practice Problem
x x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x
10 W
20 cm
5 m/s5 T
1 m 10 m