For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Forces on Current-Carrying Wires

- Moving charges in magnetic fields experience forces. Since currents in wires are moving charges, current-carrying wires in magnetic fields experience forces.
- The magnitude of the force on a current-carrying wire perpendicular to a uniform magnetic field is given by F=ILB.
- Electric motors operate by placing a current-carrying loop of wire in a uniform magnetic field, creating a net torque. As the loop of wire rotates 180 degrees, the current must either switch direction or be turned off with a device known as a commutator.

### Forces on 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
- Forces on Current-Carrying Wires 0:42
- Moving Charges in Magnetic Fields Experience Forces
- Current in a Wire is Just Flow of Charges
- 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

### AP Physics C: Electricity and Magnetism Online Course

I. Electricity | ||
---|---|---|

Electric Charge & Coulomb's Law | 30:48 | |

Electric Fields | 1:19:22 | |

Gauss's Law | 52:53 | |

Electric Potential & Electric Potential Energy | 1:14:03 | |

Electric Potential Due to Continuous Charge Distributions | 1:01:28 | |

Conductors | 20:35 | |

Capacitors | 41:23 | |

II. Current Electricity | ||

Current & Resistance | 17:59 | |

Circuits I: Series Circuits | 29:08 | |

Circuits II: Parallel Circuits | 39:09 | |

RC Circuits: Steady State | 34:03 | |

RC Circuits: Transient Analysis | 1:01:07 | |

III. Magnetism | ||

Magnets | 8:38 | |

Moving Charges In Magnetic Fields | 29:07 | |

Forces on Current-Carrying Wires | 17:52 | |

Magnetic Fields Due to Current-Carrying Wires | 24:43 | |

The Biot-Savart Law | 21:50 | |

Ampere's Law | 26:31 | |

Magnetic Flux | 7:24 | |

Faraday's Law & Lenz's Law | 1:04:33 | |

IV. Inductance, RL Circuits, and LC Circuits | ||

Inductance | 6:41 | |

RL Circuits | 42:17 | |

LC Circuits | 9:47 | |

V. Maxwell's Equations | ||

Maxwell's Equations | 3:38 | |

VI. Sample AP Exams | ||

1998 AP Practice Exam: Multiple Choice Questions | 32:33 | |

1998 AP Practice Exam: Free Response Questions | 29:55 |

### Transcription: Forces on Current-Carrying Wires

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

*I'm Dan Fullerton, and in this lesson, we are going to talk about forces on current-carrying wires.*0003

*Our objectives include calculating the magnitude and direction of the force on a straight segment*0009

*of current-carrying wire in a uniform magnetic field.*0013

*Indicating the direction of magnetic forces on a current-carrying loop of wire in a magnetic field.*0017

*Determining how the loop will tend to rotate as a consequence of those forces.*0022

*It is starting to talk about rotation and electric motors.*0026

*Finally, calculating the magnitude and direction of the torque experienced by rectangular loop of wire carrying-current and the magnetic field.*0031

*Again, electric motors.*0037

*Let us start by talking about the basics of the forces on current-carrying wires.*0041

*Moving charges in magnetic fields experience magnetic forces.*0046

*Current and wire though, is just the flow of charges.*0050

*Therefore, that wire is going to experience forces that are moving in a mode that is perpendicular to the magnetic fields.*0053

*If we look down here at this drawing, let us assume we have current I here in black flowing in a wire.*0062

*We are going to break up this wire into little tiny pieces DL, along the wire.*0069

*At any point in time, we have some small amount of charge DQ at the section of the wire,*0073

*moving with an instantaneous velocity V as shown here in green.*0078

*All of this is happening in the magnetic field, where at this point our magnetic field is here in blue B.*0082

*In order to find the magnetic force, what we are going to do is say that the magnetic force FB =QV × B.*0090

*Since we are only looking at a very little bit of charge, let us look at the tiny bit of force.*0104

*The differential of the magnetic force then will be DQ V × B.*0109

*That little bit of charge is velocity × the magnetic field at that point.*0119

*This implies then that the magnetic force is, let us take a look here.*0124

*If we think of it as current being equal to DQ DT.*0132

*DQ we can write as IDT.*0139

*The magnetic force is equal to, we will replace DQ with IDT times V × B.*0143

*But we also know that V is DL DT.*0157

*Therefore, V DT must be DL.*0167

*We can now write this as the differential of that magnetic force is equal to, we have our I.*0174

*V DT we said was going to be DL so that will be DL × B.*0185

*Or taking the integral to get the whole thing, the integral of DFB is going to be the integral of I DL × B.*0196

*The integral of the differential of the magnetic force is going to give you the entire magnetic force.*0212

*To finish this up, the magnetic force is going to be the integral of the current times DL × the magnetic field B.*0219

*That is how we can find the force of the current-carrying wire due to a perpendicular magnetic field.*0233

*As we have mentioned before, we can use the right hand rule to find the direction of that.*0243

*Same idea, point the fingers of your right hand in the direction of the velocity of positive charges,*0248

*then your fingers inward in the direction of the magnetic field.*0253

*Your thumb is going to point in the direction of the magnetic force.*0255

*Let us take a look at a simple example here.*0261

*A current through a wire, a straight wire of length 1 m carries a current of 100 amp through a magnetic field of 1 tesla.*0264

*Find the force on the wire.*0272

*We will start with our formula, where the magnetic force is equal to the integral of I DL × B.*0274

*This implies then that the magnetic force is going to be equal to,*0288

*as we do the integral over the entire length of wire DL, we are just going to have DL and*0292

*B are perpendicular so DL × B, that is sin θ vector.*0298

*Θ is going to be 90° and I end up with I LB sin θ.*0302

*If θ = 90°, sin θ = 1.*0312

*Therefore, our magnetic force is going to be equal to ILB.*0318

*We can substitute in our values because we know the current, we know are length, and we know our magnetic field strength.*0325

*The magnitude of our magnetic force is going to be our current 100 amp times our length 1 m*0335

*times our magnetic field strength 1 tesla, or 100 N.*0343

*Kind of a simplistic example but they start to how we can use these formulas.*0351

*Let us take a look at the electric motor then.*0358

*Here we have current traveling through a square loop of wire.*0361

*We are doing this in a uniform magnetic field to the left, we have dimensions A here and B here.*0367

*Let us see if we can figure out how this would work.*0376

*As I start, let us take a look and say that the magnetic force on that is going to be ILB.*0380

*If I look at this piece is not going to contribute any force because it is parallel to the magnetic field.*0387

*Same here, I only have to worry about these vertical sections.*0392

*That is going to be IA, B will be the force here and the force there.*0396

*You can figure out the direction using the right hand rule.*0403

*Point the fingers of your right hand in the direction the current is flowing because conventional current is positive charges.*0406

*Then, your fingers inward in the direction of the magnetic field and your thumb pointing in the direction of the force.*0411

*Our net torque then is going to be R × our force, which in this case our R.*0417

*If this is our axis, if this is all going to be spinning around, that is going to be B/ 2 b IA B our magnetic field strength.*0426

*We have another contribution from this side as well, B/ 2 IAB which is going to be equal to IABB.*0437

*What we have done is we have created a counterclockwise torque.*0450

*If we go and we look at this 180° later after its spun 180°, if we left is on in this way, we would have an opposing force that is going to create the exact opposite torque.*0454

*That is not going to help us with making any sort of electric motor.*0467

*What you do is as you spin it, as you go through that 180°, you can do one of two things.*0471

*You can either reverse the current, when you are over in this stage which you do by using a device known as a commutator.*0476

*Or just turn off the current for that stage like it goes through until you are back at this point again and*0483

*you turn the current back on to give you the force, to keep your motor accelerating.*0488

*Another example, force on the wire.*0495

*We have a wire situated in the uniform magnetic field carrying a current as shown toward the bottom of your screen.*0498

*What is in the direction of the force on the wire due to the magnetic field?*0504

*That looks like a right hand rule problem.*0509

*Point the finger of your right hand in the direction the current is flowing,*0511

*then in the direction of the magnetic field and you should find that that is going to come out of the screen it you.*0514

*The magnetic force FB is going to be out of the screen by the right hand rule.*0521

*Torque question, determine the direction of the net torque on the current-carrying closed circuit here.*0534

*It gives us the direction of the current flow.*0539

*Right away I can see in the bottom portion we are not going to have any force because that is parallel to the magnetic field.*0543

*The same thing here, we are anti parallel.*0548

*We are still not going to have any force.*0551

*The angle there sin 0 and sin 180 is 0.*0553

*We only have to worry about these portion.*0557

*We have got current flowing this way on the left hand side and current flowing this way on the right hand side.*0560

*Looking at the left hand side first, right hand rule, point the fingers of your right hand down with respect to the screen.*0567

*Bend them to the right of your screen and what you should find is you have a force coming toward you on that side.*0574

*Try the same thing on the right side.*0585

*Point the fingers of your right hand up with relation to the screen then in the direction of the magnetic field.*0587

*You should find that your thumb points into the plane of the screen.*0591

*The force on that side is into the screen.*0596

*That is going to create a net torque pushing up on one side down on the other to spin our loop of wire.*0599

*Another example, determine the direction of the net torque on this current-carrying close circuit.*0611

*We got current flowing this way.*0617

*We got current flowing this way.*0619

*We would not have any contribution to force here though because it is parallel to the magnetic field.*0621

*Current flowing this way and current flowing that way.*0626

*Over here on the right, no contribution from force because it is parallel to the magnetic field.*0628

*You got to worry about the top and the bottom.*0636

*We can use our right hand rules again.*0638

*At the top, point the fingers of your right hand in the direction positive current is flowing.*0640

*Bend in the direction of the magnetic field and I come up with a force out of the plane of the page of the screen.*0644

*There is our forced there.*0650

*On the other side of course, you can go through the exercise again with your right hand rule*0652

*and find that the force is pointing into the plane of the screen.*0656

*Hopefully getting pretty good at these by know.*0662

*Let us do another one and make sure we have got it.*0663

*Determine the direction of the network on the current-carrying closed circuit.*0666

*Let us draw in our current around their close circuit.*0670

*Our magnetic field is now coming out of the screen toward you.*0675

*As I look here, let us start at the bottom.*0680

*Point the fingers of my right hand in the direction of the current flow.*0683

*Bend in the direction of the magnetic field and I come up with a force that is toward the inside of the loop.*0687

*If I do the same thing over here, I come up with a force in the same direction toward the inside*0696

*and toward the top toward the inside, to over here, toward the inside.*0701

*We are not going to have any torque.*0706

*Instead, all we are getting is forces that wanting to compress this square loop of circuit.*0708

*No torque in this one.*0718

*Let us try one more example.*0722

*This one is definitely more involved than the others.*0724

*But I think it is good to see how you would do something like this.*0727

*We have a curved wire here carrying a constant current I as shown.*0730

*3 uniform magnetic field to the right.*0734

*Find the net force acting on the wire.*0737

*First thing to note is these portions that are straight are parallel to the magnetic field*0741

*so they are not going to contribute anything as far as force goes.*0745

*We only have to worry about the curved section.*0748

*Simplifies it a little bit but we still got some work to do here.*0751

*What I’m going to do is I'm going to take a second, I'm going to draw in a couple of lines to help me out here.*0755

*We will draw that so we have got our X axis and let us take a look at just a little tiny piece of our wire right there.*0762

*At that point we are going to have some current and let us call this our IDL.*0771

*I'm going to draw in at this point also, our vertical just to help us with any angles we might have.*0780

*I’m going to draw our radius in here from our little piece to the center point of our circle.*0787

*As I do that, let me define that angle right there as θ, which also is going to make that angle write θ with a little bit of geometry.*0793

*The length of this is R.*0805

*Let us write our formula then for the magnetic force.*0809

*It is going to be the integral of IDL × B.*0814

*What is DL going to be?*0822

*DL is going to be, if we have θ moving around that part of the circle, that is going to be a RD θ sin θ in the I ̂ direction.*0826

*In the X direction, because as we look over here, as we do our DL, the sin θ is going to be the opposite piece from θ.*0843

*This piece, as we break that into X and Y components + RD θ.*0854

*The adjacent side is going to be the Y component or the J ̂ component, cos θ J ̂.*0861

*Then our magnetic force is going to be the integral of I.*0872

*Our DL is going to be RD θ sin θ I ̂ + RD θ cos θ J ̂ × our magnetic field.*0878

*But notice, our magnetic field is only in the I ̂ direction, it is to the right.*0902

*Instead of writing the vector here, let us just make that the constant B its magnitude times I ̂.*0906

*We have got to go back and figure out where we are going to integrate from.*0913

*As we go from here all the way up to here, we can integrate from θ = - π/ 2.*0917

*Π / 2 for angles.*0927

*Now, we got to do our × product here.*0930

*And here, it might be useful to go back to our math review section if you do not remember × products.*0933

*But we have got our I ̂ and J ̂ been × each other.*0939

*That is going to give us the - BR cos θ D θ in the Z direction or K ̂.*0943

*The I ̂ and I ̂, that × product piece is going to be 0.*0956

*We can do our integration.*0962

*The first thing I’m going to do is pull my constants out.*0964

*Our force is going to be, I lost my I in there somehow.*0967

*Where did I put that, I should pull that out there.*0974

*Our force is going to be IBR, we can pull the negative out as well.*0977

*- IBR K ̂, the unit vector is a constant.*0983

*K ̂ integral from -π/ 2 to θ = π/ 2 of cos θ D θ.*0988

*And that is going to be equal to, we still got our constants, -BR K ̂.*1001

*The integral of cos is sin, so we have the sin of θ is evaluated from -π/ 2 to π / 2, which is going to be equal to - IBR K ̂ sin of π/ 2 1 - sin of – π/ 2 -1.*1009

*1 - -1 is 2 so we are going to get for our magnetic force is going to be -2 IBR in the Z direction.*1037

*K ̂ unit vector in the Z direction.*1051

*Definitely one of the more involved examples and the toughest part is setting that up here.*1056

*But look through it once or twice and I'm sure you will be able to follow there.*1060

*Hopefully, that gets you a good start on forces on current-carrying wires.*1064

*Thank you so much for watching www.educator.com.*1068

*We will see you soon, make it a great day.*1070

1 answer

Last reply by: Professor Dan Fullerton

Fri Apr 15, 2016 6:09 AM

Post by Xiangyu Xu on April 14, 2016

Hello Professor Dan Fullerton.

In the example VI, I wonder why cos?j^ x Bi^ gives us -cos? in k^?

0 answers

Post by Professor Dan Fullerton on January 18, 2016

Note that we have 3 right hand rules to learn about magnetism.

0 answers

Post by Professor Dan Fullerton on January 18, 2016

You're looking for the direction of a force given a moving charge direction and a magnetic field direction. What you're actually doing is determining the direction of the cross product.

0 answers

Post by Parth Shorey on January 16, 2016

Why is example II a right hand rule?