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Lecture Comments (5)

1 answer

Last reply by: Professor Dan Fullerton
Fri Apr 15, 2016 6:09 AM

Post by Xiangyu Xu on April 14 at 10:10:56 PM

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 at 07:18:54 AM

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

0 answers

Post by Professor Dan Fullerton on January 18 at 07:18:40 AM

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 at 02:43:20 PM

Why is example II a right hand rule?

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

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 and0483

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