AP Physics C: Electricity & Magnetism 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