  Dan Fullerton

Magnetic Fields Due to Current-Carrying Wires

Slide Duration:

Section 1: Electricity
Electric Charge & Coulomb's Law

30m 48s

Intro
0:00
Objective
0:15
Electric Charges
0:50
Matter is Made Up of Atoms
0:52
Most Atoms are Neutral
1:02
Ions
1:11
Coulomb
1:18
Elementary Charge
1:34
Law of Conservation of Charge
2:03
Example 1
2:39
Example 2
3:42
Conductors and Insulators
4:41
Conductors Allow Electric Charges to Move Freely
4:43
Insulators Do Not Allow Electric Charges to Move Freely
4:50
Resistivity
4:58
Charging by Conduction
5:32
Conduction
5:37
Balloon Example
5:40
Charged Conductor
6:14
Example 3
6:28
The Electroscope
7:16
Charging by Induction
7:57
Bring Positive Rod Near Electroscope
8:08
Ground the Electroscope
8:27
Sever Ground Path and Remove Positive Rod
9:07
Example 4
9:39
Polarization and Electric Dipole Moment
11:46
Polarization
11:54
Electric Dipole Moment
12:05
Coulomb's Law
12:38
Electrostatic Force, Also Known as Coulombic Force
12:48
How Force of Attraction or Repulsion Determined
12:55
Formula
13:08
Coulomb's Law: Vector Form
14:18
Example 5
16:05
Example 6
18:25
Example 7
19:14
Example 8
23:21
Electric Fields

1h 19m 22s

Intro
0:00
Objectives
0:09
Electric Fields
1:33
Property of Space That Allows a Charged Object to Feel a Force
1:40
Detect the Presence of an Electric Field
1:51
Electric Field Strength Vector
2:03
Direction of the Electric Field Vector
2:21
Example 1
3:00
Visualizing the Electric Field
4:13
Electric Field Lines
4:56
E Field Due to a Point Charge
7:19
Derived from the Definition of the Electric Field and Coulomb's Law
7:24
Finding the Electric Field Due to Multiple Point Charges
8:37
Comparing Electricity to Gravity
8:51
Force
8:54
Field Strength
9:09
Constant
9:19
Charge Units vs. Mass Units
9:35
Attracts vs. Repel
9:44
Example 2
10:06
Example 3
17:25
Example 4
24:29
Example 5
25:23
Charge Densities
26:09
Linear Charge Density
26:26
Surface Charge Density
26:30
Volume Charge Density
26:47
Example 6
27:26
Example 7
37:07
Example 8
50:13
Example 9
54:01
Example 10
1:03:10
Example 11
1:13:58
Gauss's Law

52m 53s

Intro
0:00
Objectives
0:07
Electric Flux
1:16
Amount of Electric Field Penetrating a Surface
1:19
Symbol
1:23
Point Charge Inside a Hollow Sphere
4:31
Place a Point Charge Inside a Hollow Sphere of Radius R
4:39
Determine the Flux Through the Sphere
5:09
Gauss's Law
8:39
Total Flux
8:59
Gauss's Law
9:10
Example 1
9:53
Example 2
17:28
Example 3
22:37
Example 4
25:40
Example 5
30:49
Example 6
45:06
Electric Potential & Electric Potential Energy

1h 14m 3s

Intro
0:00
Objectives
0:08
Electric Potential Energy
0:58
Gravitational Potential Energy
1:02
Electric Potential Energy
1:11
Electric Potential
1:19
Example 1
1:59
Example 2
3:08
The Electron-Volt
4:02
Electronvolt
4:16
1 eV is the Amount of Work Done in Moving an Elementary Charge Through a Potential Difference of 1 Volt
4:26
Conversion Ratio
4:41
Example 3
4:52
Equipotential Lines
5:35
Topographic Maps
5:36
Lines Connecting Points of Equal Electrical Potential
5:47
Always Cross Electrical Field Lines at Right Angles
5:57
Gradient of Potential Increases As Equipotential Lines Get Closer
6:02
Electric Field Points from High to Low Potential
6:27
Drawing Equipotential Lines
6:49
E Potential Energy Due to a Point Charge
8:20
Electric Force from Electric Potential Energy
11:59
E Potential Due to a Point Charge
13:07
Example 4
14:42
Example 5
15:59
Finding Electric Field From Electric Potential
19:06
Example 6
23:41
Example 7
25:08
Example 8
26:33
Example 9
29:01
Example 10
31:26
Example 11
43:23
Example 12
51:51
Example 13
58:12
Electric Potential Due to Continuous Charge Distributions

1h 1m 28s

Intro
0:00
Objectives
0:10
Potential Due to a Charged Ring
0:27
Potential Due to a Uniformly Charged Desk
3:38
Potential Due to a Spherical Shell of Charge
11:21
Potential Due to a Uniform Solid Sphere
14:50
Example 1
23:08
Example 2
30:43
Example 3
41:58
Example 4
51:41
Conductors

20m 35s

Intro
0:00
Objectives
0:08
Charges in a Conductor
0:32
Charge is Free to Move Until the
0:36
All Charge Resides at Surface
2:18
Field Lines are Perpendicular to Surface
2:34
Electric Field at the Surface of a Conductor
3:04
Looking at Just the Outer Surface
3:08
Large Electric Field Where You Have the Largest Charge Density
3:59
Hollow Conductors
4:22
Draw Hollow Conductor and Gaussian Surface
4:36
Applying Gaussian Law
4:53
Any Hollow Conductor Has Zero Electric Field in Its Interior
5:24
5:35
Electric Field and Potential Due to a Conducting Sphere
6:03
Example 1
7:31
Example 2
12:39
Capacitors

41m 23s

Intro
0:00
Objectives
0:08
What is a Capacitor?
0:42
Electric Device Used to Store Electrical Energy
0:44
Place Opposite Charges on Each Plate
1:10
Develop a Potential Difference Across the Plates
1:14
Energy is Stored in the Electric Field Between the Plates
1:17
Capacitance
1:22
Ratio of the Charge Separated on the Plates of a Capacitor to the Potential Difference Between the Plates
1:25
Units of Capacitance
1:32
1:37
Formula
1:52
Calculating Capacitance
1:59
Assume Charge on Each Conductor
2:05
Find the Electric Field
2:11
Calculate V by Integrating the Electric Field
2:21
Utilize C=Q/V to Solve for Capitance
2:33
Example 1
2:44
Example 2
5:30
Example 3
10:46
Energy Stored in a Capacitor
15:25
Work is Done Charging a Capacitor
15:28
Solve For That
15:55
Field Energy Density
18:09
Amount of Energy Stored Between the Plates of a Capacitor
18:11
Example
18:25
Dielectrics
20:44
Insulating Materials Place Between Plates of Capacitor to Increase The Devices' Capacitance
20:47
Electric Field is Weakened
21:00
The Greater the Amount of Polarization The Greater the Reduction in Electric Field Strength
21:58
Dielectric Constant (K)
22:30
Formula
23:00
Net Electric Field
23:35
Key Take Away Point
23:50
Example 4
24:00
Example 5
25:50
Example 6
26:50
Example 7
28:53
Example 8
30:57
Example 9
32:55
Example 10
34:59
Example 11
37:35
Example 12
39:57
Section 2: Current Electricity
Current & Resistance

17m 59s

Intro
0:00
Objectives
0:08
Electric Current
0:44
Flow Rate of Electric Charge
0:45
Amperes
0:49
Positive Current Flow
1:01
Current Formula
1:19
Drift Velocity
1:35
Constant Thermal Motion
1:39
Net Electron Flow
1:43
When Electric Field is Applied
1:49
Electron Drift Velocity
1:55
Derivation of Current Flow
2:12
Apply Electric Field E
2:20
Define N as the Volume Density of Charge Carriers
2:27
Current Density
4:33
Current Per Area
4:36
Formula
4:44
Resistance
5:14
Ratio of the Potential Drop Across an Object to the Current Flowing Through the Object
5:19
5:23
Resistance of a Wire
6:05
Depends on Resistivity
6:09
Resistivity Relates to the Ability of a Material to Resist the Flow of Electrons
6:25
Refining Ohm's Law
7:22
Conversion of Electric Energy to Thermal Energy
8:23
Example 1
9:54
Example 2
10:54
Example 3
11:26
Example 4
14:41
Example 5
15:24
Circuits I: Series Circuits

29m 8s

Intro
0:00
Objectives
0:08
Ohm's Law Revisited
0:39
Relates Resistance, Potential Difference, and Current Flow
0:39
Formula
0:44
Example 1
1:09
Example 2
1:44
Example 3
2:15
Example 4
2:56
Electrical Power
3:26
Transfer of Energy Into Different Types
3:28
Light Bulb
3:37
Television
3:41
Example 5
3:49
Example 6
4:27
Example 7
5:12
Electrical Circuits
5:42
Closed-Loop Path Which Current Can Flow
5:43
Typically Comprised of Electrical Devices
5:52
Conventional Current Flows from High Potential to Low Potential
6:04
Circuit Schematics
6:26
Three-dimensional Electrical Circuits
6:37
Source of Potential Difference Required for Current to Flow
7:29
Complete Conducting Paths
7:42
Current Only Flows in Complete Paths
7:43
Left Image
7:46
Right Image
7:56
Voltmeters
8:25
Measure the Potential Difference Between Two Points in a Circuit
8:29
Can Remove Voltmeter from Circuit Without Breaking the Circuit
8:47
Very High Resistance
8:53
Ammeters
9:31
Measure the Current Flowing Through an Element of a Circuit
9:32
Very Low Resistance
9:46
Put Ammeter in Correctly
10:00
Example 8
10:24
Example 9
11:39
Example 10
12:59
Example 11
13:16
Series Circuits
13:46
Single Current Path
13:49
Removal of Any Circuit Element Causes an Open Circuit
13:54
Kirchhoff's Laws
15:48
Utilized in Analyzing Circuits
15:54
Kirchhoff's Current Law
15:58
Junction Rule
16:02
Kirchhoff's Voltage Law
16:30
Loop Rule
16:49
Example 12
16:58
Example 13
17:32
Basic Series Circuit Analysis
18:36
Example 14
22:06
Example 15
22:29
Example 16
24:02
Example 17
26:47
Circuits II: Parallel Circuits

39m 9s

Intro
0:00
Objectives
0:16
Parallel Circuits
0:38
Multiple Current Paths
0:40
Removal of a Circuit Element May Allow Other Branches of the Circuit to Continue Operating
0:44
Draw a Simple Parallel Circuit
1:02
Basic Parallel Circuit Analysis
3:06
Example 1
5:58
Example 2
8:14
Example 3
9:05
Example 4
11:56
Combination Series-Parallel Circuits
14:08
Circuit Doesn't Have to be Completely Serial or Parallel
14:10
Look for Portions of the Circuit With Parallel Elements
14:15
Lead to Systems of Equations to Solve
14:42
Analysis of a Combination Circuit
14:51
Example 5
20:23
Batteries
28:49
Electromotive Force
28:50
Pump for Charge
29:04
Ideal Batteries Have No Resistance
29:10
Real Batteries and Internal Resistance
29:20
Terminal Voltage in Real Batteries
29:33
Ideal Battery
29:50
Real Battery
30:25
Example 6
31:10
Example 7
33:23
Example 8
35:49
Example 9
38:43

34m 3s

Intro
0:00
Objectives
0:17
Capacitors in Parallel
0:51
Store Charge on Plates
0:52
Can Be Replaced with an Equivalent Capacitor
0:56
Capacitors in Series
1:12
Must Be the Same
1:13
Can Be Replaced with an Equivalent Capacitor
1:15
RC Circuits
1:30
Comprised of a Source of Potential Difference, a Resistor Network, and Capacitor
1:31
RC Circuits from the Steady-State Perspective
1:37
Key to Understanding RC Circuit Performance
1:48
Charging an RC Circuit
2:08
Discharging an RC Circuit
6:18
The Time Constant
8:49
Time Constant
8:58
By 5 Time Constant
9:19
Example 1
9:45
Example 2
13:27
Example 3
16:35
Example 4
18:03
Example 5
19:39
Example 6
26:14
RC Circuits: Transient Analysis

1h 1m 7s

Intro
0:00
Objectives
0:13
Charging an RC Circuit
1:11
Basic RC Circuit
1:15
Graph of Current Circuit
1:29
Graph of Charge
2:17
Graph of Voltage
2:34
Mathematically Describe the Charts
2:56
Discharging an RC Circuit
13:29
Graph of Current
13:47
Graph of Charge
14:08
Graph of Voltage
14:15
Mathematically Describe the Charts
14:30
The Time Constant
20:03
Time Constant
20:04
By 5 Time Constant
20:14
Example 1
20:39
Example 2
28:53
Example 3
27:02
Example 4
44:29
Example 5
55:24
Section 3: Magnetism
Magnets

8m 38s

Intro
0:00
Objectives
0:08
Magnetism
0:35
Force Caused by Moving Charges
0:36
Dipoles
0:40
Like Poles Repel, Opposite Poles Attract
0:53
Magnetic Domains
0:58
Random Domains
1:04
Net Magnetic Field
1:26
Example 1
1:40
Magnetic Fields
2:03
Magnetic Field Strength
2:04
Magnets are Polarized
2:16
Magnetic Field Lines
2:53
Show the Direction the North Pole of a Magnet Would Tend to Point if Placed on The Field
2:54
Direction
3:25
Magnetic Flux
3:41
The Compass
4:05
Earth is a Giant Magnet
4:07
Earth's Magnetic North Pole
4:10
Compass Lines Up with the Net Magnetic Field
4:48
Magnetic Permeability
5:00
Ratio of the magnetic Field Strength Induced in a Material to the Magnetic Field Strength of the Inducing Field
5:01
Free Space
5:13
Permeability of Matter
5:41
Highly Magnetic Materials
5:47
Magnetic Dipole Moment
5:54
The Force That a Magnet Can Exert on Moving Charges
5:59
Relative Strength of a Magnet
6:04
Example 2
6:26
Example 3
6:52
Example 4
7:32
Example 5
7:57
Moving Charges In Magnetic Fields

29m 7s

Intro
0:00
Objectives
0:08
Magnetic Fields
0:57
Vector Quantity
0:59
Tesla
1:08
Gauss
1:14
Forces on Moving Charges
1:30
Magnetic Force is Always Perpendicular to the Charged Objects Velocity
1:31
Magnetic Force Formula
2:04
Magnitude of That
2:20
Image
2:29
Direction of the Magnetic Force
3:54
Right-Hand Rule
3:57
Electron of Negative Charge
4:04
Example 1
4:51
Example 2
6:58
Path of Charged Particles in B Fields
8:07
Magnetic Force Cannot Perform Work on a Moving Charge
8:08
Magnetic Force Can Change Its Direction
8:11
Total Force on a Moving Charged Particle
9:40
E Field
9:50
B Field
9:54
Lorentz Force
9:57
Velocity Selector
10:33
Charged Particle in Crosses E and B Fields Can Undergo Constant Velocity Motion
10:37
Particle Can Travel Through the Selector Without Any Deflection
10:49
Mass Spectrometer
12:21
Magnetic Fields Accelerate Moving Charges So That They Travel in a Circle
12:26
Used to Determine the Mass of An Unknown Particle
12:32
Example 3
13:11
Example 4
15:01
Example 5
16:44
Example 6
17:33
Example 7
19:12
Example 8
19:50
Example 9
24:02
Example 10
25:21
Forces on Current-Carrying Wires

17m 52s

Intro
0:00
Objectives
0:08
Forces on Current-Carrying Wires
0:42
Moving Charges in Magnetic Fields Experience Forces
0:45
Current in a Wire is Just Flow of Charges
0:49
Direction of Force Given by RHR
4:04
Example 1
4:22
Electric Motors
5:59
Example 2
8:14
Example 3
8:53
Example 4
10:09
Example 5
11:04
Example 6
12:03
Magnetic Fields Due to Current-Carrying Wires

24m 43s

Intro
0:00
Objectives
0:08
Force on a Current-Carrying Wire
0:38
Magnetic Fields Cause a Force on Moving Charges
0:40
Current Carrying Wires
0:44
How to Find the Force
0:55
Direction Given by the Right Hand Rule
1:04
Example 1
1:17
Example 2
2:26
Magnetic Field Due to a Current-Carrying Wire
4:20
Moving Charges Create Magnetic Fields
4:24
Current-Carrying Wires Carry Moving Charges
4:27
Right Hand Rule
4:32
Multiple Wires
4:51
Current-Carrying Wires Can Exert Forces Upon Each Other
4:58
First Right Hand Rule
5:15
Example 3
6:46
Force Between Parallel Current Carrying Wires
8:01
Right Hand Rules to Determine Force Between Parallel Current Carrying Wires
8:03
Find Magnetic Field Due to First Wire, Then Find Direction of Force on 2nd Wire
8:08
Example
8:20
Gauss's Law for Magnetism
9:26
Example 4
10:35
Example 5
12:57
Example 6
14:19
Example 7
16:50
Example 8
18:15
Example 9
18:43
The Biot-Savart Law

21m 50s

Intro
0:00
Objectives
0:07
Biot-Savart Law
0:24
Brute Force Method
0:49
Draw It Out
0:54
Diagram
1:35
Example 1
3:43
Example 2
7:02
Example 3
14:31
Ampere's Law

26m 31s

Intro
0:00
Objectives
0:07
Ampere's Law
0:27
Finds the Magnetic Field Due to Current Flowing in a Wire in Situations of Planar and Cylindrical Symmetry
0:30
Formula
0:40
Example
1:00
Example 1
2:19
Example 2
4:08
Example 3
6:23
Example 4
8:06
Example 5
11:43
Example 6
13:40
Example 7
17:54
Magnetic Flux

7m 24s

Intro
0:00
Objectives
0:07
Magnetic Flux
0:31
Amount of Magnetic Field Penetrating a Surface
0:32
Webers
0:42
Flux
1:07
Total Magnetic Flux
1:27
Magnetic Flux Through Closed Surfaces
1:51
Gauss's Law for Magnetism
2:20
Total Flux Magnetic Flux Through Any Closed Surface is Zero
2:23
Formula
2:45
Example 1
3:02
Example 2
4:26

1h 4m 33s

Intro
0:00
Objectives
0:08
0:44
0:46
Direction of the Induced Current is Given by Lenz's Law
1:09
Formula
1:15
Lenz's Law
1:49
Lenz's Law
2:14
Lenz's Law
2:16
Example
2:30
Applying Lenz's Law
4:09
If B is Increasing
4:13
If B is Decreasing
4:30
Maxwell's Equations
4:55
Gauss's Law
4:59
Gauss's Law for Magnetism
5:16
Ampere's Law
5:26
5:39
Example 1
6:14
Example 2
9:36
Example 3
11:12
Example 4
19:33
Example 5
26:06
Example 6
31:55
Example 7
42:32
Example 8
48:08
Example 9
55:50
Section 4: Inductance, RL Circuits, and LC Circuits
Inductance

6m 41s

Intro
0:00
Objectives
0:08
Self Inductance
0:25
Ability of a Circuit to Oppose the Magnetic Flux That is Produced by the Circuit Itself
0:27
Changing Magnetic Field Creates an Induced EMF That Fights the Change
0:37
Henrys
0:44
Function of the Circuit's Geometry
0:53
Calculating Self Inductance
1:10
Example 1
3:40
Example 2
5:23
RL Circuits

42m 17s

Intro
0:00
Objectives
0:11
Inductors in Circuits
0:49
Inductor Opposes Current Flow and Acts Like an Open Circuit When Circuit is First Turned On
0:52
Inductor Keeps Current Going and Acts as a Short
1:04
If the Battery is Removed After a Long Time
1:16
Resister Dissipates Power, Current Will Decay
1:36
Current in RL Circuits
2:00
Define the Diagram
2:03
Mathematically Solve
3:07
Voltage in RL Circuits
7:51
Voltage Formula
7:52
Solve
8:17
Rate of Change of Current in RL Circuits
9:42
Current and Voltage Graphs
10:54
Current Graph
10:57
Voltage Graph
11:34
Example 1
12:25
Example 2
23:44
Example 3
34:44
LC Circuits

9m 47s

Intro
0:00
Objectives
0:08
LC Circuits
0:30
Assume Capacitor is Fully Charged When Circuit is First Turned On
0:38
Interplay of Capacitor and Inductor Creates an Oscillating System
0:42
Charge in LC Circuit
0:57
Current and Potential in LC Circuits
7:14
Graphs of LC Circuits
8:27
Section 5: Maxwell's Equations
Maxwell's Equations

3m 38s

Intro
0:00
Objectives
0:07
Maxwell's Equations
0:19
Gauss's Law
0:20
Gauss's Law for Magnetism
0:44
1:00
Ampere's Law
1:18
Revising Ampere's Law
1:49
Allows Us to Calculate the Magnetic Field Due to an Electric Current
1:50
Changing Electric Field Produces a Magnetic Field
1:58
Conduction Current
2:33
Displacement Current
2:44
Maxwell's Equations (Complete)
2:58
Section 6: Sample AP Exams
1998 AP Practice Exam: Multiple Choice Questions

32m 33s

Intro
0:00
0:11
Multiple Choice 36
0:36
Multiple Choice 37
2:07
Multiple Choice 38
2:53
Multiple Choice 39
3:32
Multiple Choice 40
4:37
Multiple Choice 41
4:43
Multiple Choice 42
5:22
Multiple Choice 43
6:00
Multiple Choice 44
8:09
Multiple Choice 45
8:27
Multiple Choice 46
9:03
Multiple Choice 47
9:30
Multiple Choice 48
10:19
Multiple Choice 49
10:47
Multiple Choice 50
12:25
Multiple Choice 51
13:10
Multiple Choice 52
15:06
Multiple Choice 53
16:01
Multiple Choice 54
16:44
Multiple Choice 55
17:10
Multiple Choice 56
19:08
Multiple Choice 57
20:39
Multiple Choice 58
22:24
Multiple Choice 59
22:52
Multiple Choice 60
23:34
Multiple Choice 61
24:09
Multiple Choice 62
24:40
Multiple Choice 63
25:06
Multiple Choice 64
26:07
Multiple Choice 65
27:26
Multiple Choice 66
28:32
Multiple Choice 67
29:14
Multiple Choice 68
29:41
Multiple Choice 69
31:23
Multiple Choice 70
31:49
1998 AP Practice Exam: Free Response Questions

29m 55s

Intro
0:00
0:14
Free Response 1
0:22
Free Response 2
10:04
Free Response 3
16:22
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• ## Related Books 4 answers Last reply by: Professor Dan FullertonTue Apr 10, 2018 7:25 AMPost by Kevin Fleming on April 9, 2018Prof. Fullerton,For example 9, part f, how we can we just use the velocity selector equation (only applicable when there is no deflection) to determine the additional electric field (which was present during deflection)? Thanks. 1 answer Last reply by: Professor Dan FullertonMon Jan 18, 2016 7:17 AMPost by Parth Shorey on January 17, 2016Why are we always using the right hand rule, and not the left hand rule?

### Magnetic Fields Due to Current-Carrying Wires

• Moving charges (including electrical currents) create magnetic fields.
• Magnetic fields exert forces on moving charges (including electrical currents).
• Right-hand rules allow you to determine the direction of forces and fields due to magnetic interactions.
• The force on a current-carrying wire in a magnetic field is given by: F=ILBsinθ.
• The magnetic field due to a current-carrying wire is given by: B=µ0*I/(2πr).
• A coil of wire known as a solenoid can be used to create an electromagnet. Placing an iron core inside a solenoid greatly strengthens the electromagnet.
• Current-carrying parallel wires with current in the same direction attract each other; if the current flows in opposite directions they repel each other.

### Magnetic Fields Due to Current-Carrying Wires

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Objectives 0:08
• Force on a Current-Carrying Wire 0:38
• Magnetic Fields Cause a Force on Moving Charges
• Current Carrying Wires
• How to Find the Force
• Direction Given by the Right Hand Rule
• Example 1 1:17
• Example 2 2:26
• Magnetic Field Due to a Current-Carrying Wire 4:20
• Moving Charges Create Magnetic Fields
• Current-Carrying Wires Carry Moving Charges
• Right Hand Rule
• Multiple Wires
• Current-Carrying Wires Can Exert Forces Upon Each Other
• First Right Hand Rule
• Example 3 6:46
• Force Between Parallel Current Carrying Wires 8:01
• Right Hand Rules to Determine Force Between Parallel Current Carrying Wires
• Find Magnetic Field Due to First Wire, Then Find Direction of Force on 2nd Wire
• Example
• Gauss's Law for Magnetism 9:26
• Example 4 10:35
• Example 5 12:57
• Example 6 14:19
• Example 7 16:50
• Example 8 18:15
• Example 9 18:43

### Transcription: Magnetic Fields Due to Current-Carrying Wires

Hello, everyone, and welcome back to www.educator.com.0000

In this lesson, we are going to talk about magnetic fields due to current carrying wires.0003

Our objectives include calculating the force on current carrying wires in a uniform magnetic field.0008

Indicating the direction of magnetic forces on current carrying loops of wire in magnetic field.0014

Calculating the magnitude and direction of the magnetic field in a point in the vicinity of a long current carrying wire.0019

Using super position to determine the magnetic field produced by 2 long current carrying wires.0027

Finally, calculating the force of attraction repulsion between 2 current carrying wires.0032

Magnetic fields cause a force on moving charges, we have talked about that, in current carrying wires contain moving charges.0040

Therefore, current carrying wire in a magnetic field may experience a magnetic force.0046

In our last lesson, we showed how we could find that force.0051

If it is a straight wire in a constant magnetic field, we have that that force is IL × B or ILB sin θ.0056

And the direction is given by the right hand rule.0064

When you point your fingers in the direction of positive charge movement,0066

then the fingers in the direction of the magnetic field and your thumb point in the direction of the force.0070

Let us do an example.0078

We have got a 10 m wire carrying 10 amps of current through a 5 tesla magnetic field as shown.0080

Find the magnitude and direction of the magnetic force on the wire.0086

Let us find the direction first.0090

If we have our right hand, it is positive charges point in the direction current is flowing, to the right of the screen.0092

Bend your fingers into the screen, inward into the screen and your thumb should point up indicating0098

the direction of the magnetic force on the wire.0106

Now to find its magnitude, we got a nice straight wire constant magnetic field, we can go with the magnetic force is ILB sin θ.0110

Our current is 10 amps, our length is 10 m, our magnetic field strength is 5 tesla and the sin θ sin 90° is going to be 1, that is just 100 × 5 or 500 N.0124

We could also use this to find the torque on a loop of wire like we have done before.0146

A loop of wire carrying current I is placed in a magnetic field, determine the net torque0151

around the axis of rotation right there due to the current on the wire.0155

Let us take a look at the different pieces here.0161

I'm going to call this portion 1 and will call this portion 2.0164

These portions that are parallel to the magnetic field are not going to contribute any force.0169

Force 1 is going to be ILB, or in this case our length is just H so we will call that IHB.0174

F2 of course must be the same, although the directions will be different.0184

As we figure out the directions, I point the fingers of my right hand in the direction current is flowing so toward the bottom of the screen here.0191

Then, in the direction of the magnetic field, toward the right of the screen and I find that the force / here should be directed out of the page.0198

Doing the same thing on the right hand side, I get a force directed into the page.0206

To find our torque, the torque due to the portion 1 is F × R or FR sin θ, which in this case our force is IHB.0212

Our R is going to be this L / 2, the distance from the axis of rotation to where that force is applied on the wire, that will be L/ 2.0225

Torque 2 due to this portion of the wire is going to be the same, IHB L/ 2.0235

Therefore, the total ne2rk, we add up torque 1 and 2 to get IHB L.0244

The magnetic field due to a current carrying wire.0262

Remember moving charges create magnetic fields in current carrying wires carrying charges.0265

Therefore, they create magnetic fields themselves.0270

The direction is given by the right hand rule again but it is a different right hand rule.0272

The direction, point you fingers, your thumb in the direction current is flowing and0278

wrap the fingers of your right hand around the object and you will get the radial direction of the magnetic field.0283

For multiple wires, you can add up the magnetic field from each of the wires to get the total using super position.0291

Note that the magnetic fields may interact with other moving charges so current carrying wires create0298

magnetic fields but they also interactive magnetic field so you can insert forces upon each other0303

when you have multiple current carrying wires in the system.0308

Let us take a look at the first right hand rule.0314

Pretend you are holding the wire with your right hand as if this is the wire, with the thumb in the direction of positive current flow.0317

Your fingers are wrapped in the direction of the magnetic field around it.0323

If we have a current flowing this way for example, down in the bottom left picture.0328

Point the fingers of your right hand toward the right of the screen and hold a pencil or pen, something like,0334

then your fingers are going to wrap around it.0338

Up above the wire, you are going to have the magnetic field coming out toward you as your fingers face you.0341

Down below it, they are going to be going away from you.0347

If you are to look from top down, if the current is coming out at you, the wires coming out of you.0351

Wrap the fingers of your right hand around it and you can see the direction the magnetic field or away from you, into the plane of the screen.0359

If you look at it from the side view, we have our current in 1 direction the magnetic field is radially around it.0366

Right hand rule help to figure out which direction that is going to occur.0375

Now to find the magnetic fields strength due to the current carrying wire, that is going to be μ₀ or0380

permeability × I the current / 2 π × the distance you are from the wire R.0387

Μ₀ I/ 2 π R or again μ₀ our permeability of free space is 4 π × 10⁻⁷ tesla m/ ampere.0393

Let us put this into play, a wire carries a current of 6 amps to the left.0404

Find the magnetic field at point P located 0.1 m below the wire, down here.0410

As I look at this, I’m first going to figure out the direction.0416

I’m going to do my right hand rule.0419

Point the thumb of my right hand in the direction current is flowing, toward the left of your screen.0422

Wrap the fingers of your right hand around it and up above the wire,0429

I see that my fingers are headed into the plane of the screen so the magnetic field that is created must be that direction.0432

Down below the wire, they are coming out toward me.0438

That must be toward you down below the wire so there is our direction.0442

To find its magnitude, B equals μ₀ I/ 2 π R, the magnetic field due do a long length of wire0447

which will be 4 π × 10⁻⁷ tesla m/ amp × our current 6 amps/ 2 π × the distance from our wire 0.1 m,0456

which gives us a magnetic field strength of about 1.2 × 10⁻⁵ tesla.0470

Let us talk about multiple wires.0481

You can use the right hand rule to determine the force between parallel current carrying wires.0483

Find the magnetic field due to 1 wire, draw it and then find the direction of the force on the second wire due to the current in that wire.0487

You know you must have the equal and opposite force on the other wire due to Newton’s third law.0494

You have 2 wires in the same direction, they attract each other.0500

Let us walk through that.0503

Let us first take a look at this bottom wire and find the magnetic field due to that bottom wire.0505

As I look at this using the right hand rule, I can see that up above the wire0511

I'm going to have magnetic field coming out of the plane of the page and down below it, it is into it.0514

We have got magnetic field coming out due to do just this bottom wire.0521

Now we can use the right hand rule on charges moving through a magnetic field on our top wire.0528

Our positive charges are moving to the right as you point the fingers of your right hand to the right of your screen,0533

bend them in the direction of the magnetic field out of the plane of the screen, you are going to see that you get a force on the top wire toward the bottom.0540

You can do the same thing to figure out that for the bottom wire is going to feel a force toward the top.0551

They attract each other.0555

I will leave it to you to try that to prove that the wires repel each other when the currents are going in opposite directions.0557

Here is a great time to talk about Gauss’s law for magnetism.0567

If you remember for electricity, how useful Gauss’s law was for finding the electric field of symmetric charge distributions.0570

We have a similar law for magnetism, you can never draw a close surface with any net magnetic flux0578

because there are no magnetic monopoles.0585

Whatever flux goes into a close surface also has to leave.0588

That is the basis of Gauss’s law for magnetism.0594

The total net magnetic flux is the integral over the close surface of E dot DA always has to be 0,0596

because you cannot have a north without a south.0604

If you can have a just and north, this can be something other than 0.0606

But because north always come with south, whatever flux lines go into a close surface also come out.0610

The net has to be 0.0616

This is another of our Maxwell’s equations that form the backbone of the course.0618

Our first was Gauss’s law for electricity.0623

Here is our second Gauss’s law for magnetism and that will leave us 2 more, Faraday’s law and amperes law which are coming up shortly.0625

Let us take a look at fields due to wires.0635

2 long current carrying wires are separated by a distance D as shown, what is the net magnetic field0638

due to these wires at point P located halfway between the 2 wires that the top wire carries a current of 3 amps0645

and the bottom wire carries a current of 5 amps?0651

I would use super position to do this.0655

Let us call our top wire and we call this wire A, and we will call our bottom wire, wire B.0658

We will first figure out the magnetic field at P due to A and I will add it to the magnetic field at P due to B.0666

The magnetic field at P due to A, we can find using μ₀ I/ 2 π R which is μ₀ × 3 amps/ 2 π × D/ 2,0673

which is just going to be 3 μ₀/ π D.0690

The direction of that by the right hand rule is going to be into the plane of the screen.0696

Let us see what we get for the magnetic field at P due to B.0703

That is going to be, μ₀ I/ 2 π R, which is μ₀ × 5 amps/ 2 π × D/ 2, or 5 μ₀/ π D.0711

The direction now is going to be into the plane of the screen.0729

If we want the total magnetic field at point P, we had up our 2 contributions from A and B.0733

It is going to be B at point P due to A + B at point P2 due to wire B,0744

which is just going to be 3 μ₀/ π D + 5 μ₀/ π D is going to be 8 μ₀/ π D, into the plane of the screen.0755

Very good, let us take a look at the next problem.0769

Determine whether the 2 wires will feel an attractive magnetic force0778

or repulsive magnetic force due to the currents that they carry in opposite directions, which will experience a stronger force?0781

We can do this just like we did previously.0789

First off, find the magnetic field due to one of the wires and then see how it interacts with the current flowing through the other wire.0792

If we start with the top this time, let us take a look at its magnetic field because of current traveling to the right,0798

using the right hand rule I can see that we will have magnetic field coming out0806

of the plane of the screen above the wire and into the plane of the screen down below the wire.0812

Now we use the right hand rule down below and find out that we are going to have a force0819

to the bottom on the bottom wire due to the magnetic field caused by the top.0825

Newton’s third law says you must have the same magnitude opposite direction force up there.0831

We are going to have repulsive forces, when you are going in opposite directions.0837

Which one will experience the stronger force though?0844

That is a trick question.0846

By Newton’s third law, the magnitude of those forces says it is going to be the same.0848

Opposite directions but same magnitude.0853

Two wires carry a current as shown below, find the magnetic field at point P right there.0860

Let us call this I1 and maybe call this I2, just to help clarify as we do things.0866

We can do this by super position again.0873

Let us start off by looking at the magnetic field due to wire 1 which will be μ₀ I/ 2 π R or 4 π × 10⁻⁷0876

× 4 amps/ 2 π × our R 0.15 m, which I will get is about 5.33 × 10⁻⁶ tesla.0892

The direction by the right hand rule is going to be into the plane of the page.0905

Same basic idea for the magnetic field from wire 2.0911

That is going to be μ₀ I/ 2 π R or 4 π × 10⁻⁷ 3 amps in this wire/ 2 π × 0.1 m.0916

I come up with about 6 × 10⁻⁶ tesla and this one at point P is going to be coming out of the plane of the screen.0933

When we do the totals now, we start writing it over here so we have a little bit more room.0943

B at point P total is going to be B due to 1 + B due to 2, which is going to be,0949

we got to pick a direction to call positive because these are in opposite directions now.0964

I'm going to call out of the plane of the screen positive and into the plane of the screen negative.0968

Let us say that this is, B1 will be -5.33 × 10⁻⁶ tesla + 6 × 10⁻⁶ tesla out of the screen.0975

We are going to come up with a total of about 6.7 × 10⁻⁷ tesla out of the plane of the screen.0988

Out of screen, pretty straightforward.1002

Thanks to super position again.1008

A 5 m long straight wire runs in the 45° angle to a uniform magnetic field of 5 tesla.1012

If the force on the wire is 1 N, determine the current in the wire.1019

Let us see, our length is 5 m, our θ is 45°, our magnetic field strength is 5 tesla, our force is 1 N.1025

We are looking for current.1044

A long straight wire looks like this is a pretty straightforward problem.1047

Force is going to be ILB sin θ which implies then that our current is going to be F, our force/ LB sin θ which is our 1 N/ 5 m 5 tesla sin of 45°.1052

Therefore, our current I must be 0.0566 amp or about 56.6 milli amps.1077

Let us take a look at some more wires.1095

Determine the direction of the force on the 2 current carrying wires as shown below.1097

I think we have done this by now.1100

Hopefully by now, you have recognized that if we have currents traveling in opposite directions, they are going to repel.1103

Once you got that down, currents in opposite directions are going to repel the wires.1109

In the same direction, they are going to attract.1113

Let us finish up by looking at an old AP problem , we will take the 2009 APC E and M test free response question number 2.1120

The link to where you can find these questions is here or you can google it.1131

Take a minute, look at the question, print it out, give it a try.1135

Then, we will come back and see if we can do it together here.1139

Here we have a rectangular bar that has a 9 V battery hooked to it and it gives us the area and resistivity,1149

says electrons are the sole charge carriers in the bar.1157

The wires have negligible resistance and the switch is closed at time T equal 0.1160

Find the power delivered to the circuit by the battery.1165

For part A, power is current × voltage or V²/ R.1169

We know V but we do not know R yet but we can calculate R because R = resistivity × length/ × sectional area.1179

It tells us our resistivity is 4.5 × 10⁻⁴ ohm meters.1189

Our length is 0.08 m and the × sectional area is 5 × 10⁻⁶ m².1198

I come up with a resistance of about 7.2 ohms.1211

Now we can go back to our power equation.1219

Power is going to be equal to our potential difference 9 V² ÷ our resistance 7.2 ohms, to give us a power of 11.25 W.1221

Part A, moving on to B.1242

On the diagram, indicate the direction of the electric field in the bar.1248

That should be pretty straightforward.1253

Electric fields run from high to low potentials.1255

In the bar, we are just going to go to the right.1257

Taking a look at C then.1264

Calculate the strength of the electric field in the bar.1267

By symmetry, that should be uniform throughout the bar.1271

It is a uniform material.1273

The strength of our electric field is just going to be the potential difference ÷ the distance which is going to be 9 V / 0.08 m or about 112.5 V / m.1276

Let us take a look at part D, we have a uniform magnetic field of magnitude ¼ tesla1295

perpendicular to the bar and we want to know the magnetic force on the bar.1304

Force is ILB, the current is going to be, we know voltage, we know resistance that is going to be I equals V/ R by Ohm’s law LB,1310

which is just going to be 9 V/ 7.2 ohms × our length 0.08 m × our magnetic field strength ¼ tesla.1323

I get a force of about 0.025 N.1338

We are plugging away.1347

Part E says the electrons moving through the bar are initially deflected by the external magnetic field.1351

That only makes sense.1358

On the diagram, indicate the direction of the additional electric field that is created in the bar by the deflected electrons.1361

By the right hand rule, as those electrons are moving in, we have got to the electrons moving one way.1368

The electrons are opposite direction, current flows are going from low to high potential.1376

Since they are negative charges, left hand in the direction they are moving through the bar1382

then your fingers in the direction of the magnetic field and you see that they are moving toward the top of the bar given the lower potential.1387

Therefore, inside the bar I would expect that we would see that the additional electric field pointing that way.1394

Part F, the electrons eventually experience no deflection and move through the bar in average speed of 3.5 × 10⁻³ m/ s.1408

Find the strength of the additional electric field indicated.1418

If they experience no deflection, the electric field in the magnetic field or the electric force and1422

the magnetic force must be balanced, just like when we have our velocity selector.1428

That means that our electric force must equal our magnetic force or QE must equal Q VB,1433

or solving for the electric fields that is going to be VB which is our velocity 3.5 × 10⁻³ m/ s × our magnetic field strength 1/4 tesla.1445

And I come up with an electric field strength of about 8.75 × 10⁻⁴ V / m.1459

Hopefully that gets you a good start on fields due to the current carrying wires.1475

Thank you so much for watching www.educator.com.1478

We will see you soon, make it a great day everyone.1481

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