Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of AP Physics 1 & 2
  • Discussion

  • Study Guides

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Bookmark and Share
Lecture Comments (15)

1 answer

Last reply by: Professor Dan Fullerton
Thu Apr 14, 2016 11:51 AM

Post by Sarmad Khokhar on April 14 at 10:15:23 AM

How did you find direction of magnetic field in 6:43

1 answer

Last reply by: Professor Dan Fullerton
Thu Apr 14, 2016 11:51 AM

Post by Sarmad Khokhar on April 14 at 09:48:39 AM

How did you get length 1 meter in 5:38

1 answer

Last reply by: Professor Dan Fullerton
Thu Feb 19, 2015 8:32 PM

Post by Vibha Pandurangi on February 19, 2015

For the example with the U-Shaped wire in the freshwater (example 2), how would you determine the length of the wire?

1 answer

Last reply by: Professor Dan Fullerton
Fri Jun 27, 2014 8:46 PM

Post by Lalit Shorey on June 27, 2014

When it comes to parallel wires, why is it that one wire has the charges

1 answer

Last reply by: Professor Dan Fullerton
Thu Mar 20, 2014 5:11 PM

Post by KUNJ KASHYAP on March 20, 2014

Professor where can I find more problems like the ones shown?

4 answers

Last reply by: help me
Thu May 9, 2013 6:12 PM

Post by help me on May 8, 2013

I don't understand how to differentiate between choosing the force in or out of the page, could you please explain?

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.

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 0:09
  • Force on a Current-Carrying Wire 0:30
    • A Current-Carrying Wire in a Magnetic Field May Experience a Magnetic Force
    • Direction Given by the Right-Hand Rule
  • Example 1: Force on a Current-Carrying Wire 1:38
  • Example 2: Equilibrium on a Submerged Wire 2:33
  • Example 3: Torque on a Loop of Wire 5:55
  • Magnetic Field Due to a Current-Carrying Wire 8:49
    • Moving Charges Create Magnetic Fields
    • Wires Carry Moving Charges
    • Direction Given by the Right-Hand Rule
  • Example 4: Magnetic Field Due to a Wire 10:56
  • Magnetic Field Due to a Solenoid 12:12
    • Solenoid is a Coil of Wire
    • Direction Given by the Right-Hand Rule
  • Forces on 2 Parallel Wires 13:34
    • Current Flowing in the Same Direction
    • Current Flowing in Opposite Directions
  • Example 5: Magnetic Field Due to Wires 15:19
  • Example 6: Strength of an Electromagnet 18:35
  • Example 7: Force on a Wire 19:30
  • Example 8: Force Between Parallel Wires 20:47

Transcription: Current-Carrying Wires

Hi everyone. Thrilled to have you back with us at Educator.com. 0000

In this lesson we are going to go a little bit further with magnetism by talking about current carrying wires. 0004

Now our objectives are going to be to calculate the force on a current carrying wire in a uniform magnetic field, to indicate the direction of magnetic forces on a current carrying loop of wire in a magnetic field...0009

...to calculate the magnetic field produced by a current carrying wire, and to calculate the force of attraction or repulsion between two current carrying wires. 0020

Let us start with the basic force on a current carrying wire. 0029

We know magnetic fields cause a force on moving charges and current carrying wires are really just moving charges in a wire, so if they are in a magnetic field, there must be a force on them and we can calculate that force. 0033

That force is the current times the wire, the wire vector crossed with the magnetic field, or since we are not really dealing with cross-products too much yet...0045

...we can find the magnitude of that force on the wire is equal to the current times the length of the wire times the magnetic field strength times the sine of the angle between the wire and the magnetic field. 0054

We will find the direction of the force again using a right-hand rule. 0066

The right-hand rule we are going to use for this is extremely similar to the one we have already learned: point the fingers of your right hand in the direction of a positive charge would move or in this case, current flow. 0070

You will bend them inward toward the magnetic field; your thumb is going to point in the direction of the force, so really it is the same rule as before, just remember that the velocity of a positive particle is the same as the direction of current flow. 0081

Here we have a 10 m wire carrying 10 A of current through a 5 T magnetic field directed into the page. 0098

The (x)'s indicate the magnetic field pointing into the page. 0105

Determine the magnitude and direction of the magnetic force on the wire. 0108

Well let us find the direction first. 0112

By the right-hand rule, if current is going to the right along the page, we need to point the fingers of our right hand in that direction, then bend them 90 degrees toward that magnetic field and your thumb should point up, so the magnetic force on that wire should point up. 0114

To find its magnitude, that is going to be ILB sin(θ), where our current is 10 A, the length of our wire is 10 m, the magnetic field strength is 5 T, and our angle is 90 degrees, so sin(90) = 1, so that is going to be 500. 0131

Let us take a look here at a more detailed problem and it will bring us back to some fluids review as well. 0154

A U-shaped wire is placed in a tube and submerged in freshwater where there exists an 8 T magnetic field into the plane of the page as shown. 0160

If the mass of the tube and wire is 1 kg and the volume of the submerged tube is .003 m3, determine the current required to keep the system in static equilibrium. 0168

We want that wire to stay right where it is; it is going to have a few forces acting on it. 0178

First thing is let us figure out the direction of the magnetic force due to these different sections of wire. 0184

Over here on the right hand side, as our current is going down -- if I point the fingers of my right hand in that direction and bend them in the direction of the magnetic field, I get a magnetic force to the right. 0188

If I do the same thing over here on this section of wire, I get a magnetic force to the left. 0201

Now because we have the same current through both sections of wire and they are the same length in the uniform magnetic field, those two are going to cancel out -- force to the right, force to the left is exactly equal, so we could ignore them. 0208

This section down here that is submerged though, that is going to have a magnetic field. 0219

I point the fingers of my right hand in the direction that that is going, bend them in the direction of the magnetic field, and I see that I am going to have a magnetic force on that wire that is down. 0225

When I draw my free-body diagram (FBD) for this situation, there is my wire; I have gravity pulling down the weight; I have the magnetic force and I am going to write that as FB, but I also now have a buoyant force since it is submerged. 0236

So I do not screw things up by writing FB again and getting confused, I am just going to write that out as F(buoyant) up. 0252

They almost match up for it to stay in equilibrium, for it to stay in the exact same position. 0260

Let us do our calculations now. 0266

The buoyant force must be equal to the weight plus the magnetic force on that wire and the buoyant force, if you recall, is the density of our fluid times the volume of the fluid displaced times the acceleration due to gravity...0269

...and that must be equal to (mg) plus the magnetic force (ILB sin(θ)), and in this case, sin(θ) again is going to be sin(90 degrees) = 1. 0288

The current -- we will solve for that -- (I) must be ρvg - mg/LB.0299

So now I can substitute in my values to find that my current is going to be our density of freshwater (1,000 kg/m3), the volume displaced (.003 m3)...0310

...the acceleration due to gravity (10 m/s2) - mass (1 kg) × g (10 m/s2)/length (1 m) × magnetic field strength (8 T). 0325

If I plug that into my calculator, I come up with a 2.5 A required to hold that in position in a static equilibrium state. 0343

Example 3 is torque on a loop of wire.0356

A loop of wire carrying current (I) is placed in a magnetic field. 0359

Determine the net torque around the axis of rotation here due to the current in the wire. 0363

Well we have some different sections of wire here and just like before we can see that as we look at this in the magnetic field, the sections are going to make a difference here. 0368

Let us call this section 1; we will call this section 2, and as we look at the current as we are going to the left here toward the magnetic field, we are going in the opposite direction of the magnetic field, and the current is in the exact opposite direction. 0380

We are not going to have any force. 0396

Here as the current is traveling down, right hand in that direction, bend 90 degrees toward the magnetic field, I am going to get a force that is coming out of the plane of the page. 0397

In this section at the bottom, we are lined up with the magnetic field. 0410

We are not crossing it, so we cannot bend our fingers in the direction of the magnetic field or you could say the sine of the angle is 0, sin(θ), sin(0) is 0, so we get no magnetic force. 0413

But over here on the right, now we have current going up, we bend our fingers in the direction of the magnetic field and I get a force that is going into the plane of the page, so there. 0424

What that is going to do is create a net torque; we are going out of the plane of the page on one side and into the plane of the page on the other, so let us find the magnitude of that. 0433

The force on this section of wire -- section 1 and we will call that F1, the magnetic force - is going to be ILB sin(θ) and θ is 90 again, so sin(90) = 1...0443

...so that is going to be in this case I × H (height) × B (magnetic field strength). 0454

F2 is going to have the same magnitude; it is just coming out of the same page, so that is IHB. 0461

Now to find our torque. 0468

Torque due to this section 1 -- well that is force times the distance from our axis of rotation times the sine of the angle between those and in this case our force is IHB, and our distance -- well if that whole thing is L, this is at L/2 from that axis of rotation. 0470

Our torque from section 2 of the wire over here, also again has the same magnitude, IHB, L/2, so our net torque...0491

...which is just torque 1 + torque 2 or IHBL/2 + IHBL/2 is just going to be IHB × L. 0504

By running current through this loop of wire in a constant magnetic field, you can create a torque that is going to cause it to rotate -- electric motor. 0518

Magnetic field due to current carrying wire -- moving charges create magnetic fields and wires carry moving charges, so you must have a magnetic field around a current carrying wire. 0530

The magnitude of that magnetic field created by a current carrying wire is given by this formula -- B = μ0, the permeability of free space, times the current flow in amps divided by 2π times the distance from the center of that wire. 0542

The direction again is given by a right-hand rule, but it is another right-hand rule now. 0558

Point your thumb in the direction of the current flow. 0563

Let us assume this is our wire. 0568

If I point my thumb in the direction of current flow -- if the current is flowing up -- as I wrap my hand around the wire, that is the direction of the magnetic field. 0569

I am going to get a circular magnetic field around the wire. 0578

It is stronger closer to the wire and weaker the further away you get related to 1/r and the direction, whether it is clockwise or counter-clockwise, you use that right-hand rule -- point your thumb in the direction of positive current flow and wrap your fingers around. 0582

That is going to show you the direction of the current flow around that wire. 0597

For example -- down here in the bottom left -- if the current is coming out of the plane of the page or out of the screen toward you, point your thumb in that direction, wrap your fingers around it and you should see that you get a counter-clockwise rotation as your fingers move forward. 0602

Going into the plane of the page or into the plane of the screen however -- the current going away from you -- wrap your fingers that way and you should see a clockwise direction for the magnetic field. 0617

Over here on the right, we are just showing that same thing from a side view. 0630

If the current is going sideways, point the thumb of your right hand in the direction of the positive current flow above the wire and it is going to be coming out toward you and under the wire it is going to be going away from you; it is wrapping around that way. 0633

It is kind of tough to visualize here on screen, but try and do it yourself and see if you can start to picture these in three dimensions. 0647

A wire carries a current of 6 A to the left. Find the magnetic field at point (P) located at .1 m below the wire. 0656

As we do this, first thing we want to do is figure out the direction. 0667

If we have current flowing that direction toward the left, as I wrap the fingers of my right hand around it, I see that up above the wire, it is going to be going into the plane of the page and below it it is going to be coming out toward me. 0672

So the direction of the magnetic field is going to be toward us here at point (P). 0688

Its magnitude is given by B = μ0I/2πr, where μ0 = 4π × 10-7, the permeability of free space times our current (6 A) divided by 2π...0693

...and (r) the distance from the center of a wire to our point, which was .1 m. 0711

That is going to give me a magnetic field strength of 1.2 × 10-5 T. 0719

Well, you can make an even stronger magnetic field by making a coil of wire. 0732

As you coil that wire -- we call that a solenoid, that can be used to create a very strong magnetic field, an electromagnet. 0736

The magnetic field due to a solenoid is equal to μ0 × N (the number of loops per length) × the current flow through your wire and that (N), the number of loops per length is just going to be -- we will call that N/L, where (N) is the total number of loops and (L) is going to be the length of your solenoid. 0743

Again, the direction is given by a right-hand rule. 0767

Wrap the fingers in the direction of current flow around the solenoid and your thumb will point to the North Pole of your electromagnet. 0771

For example, if I have the wire wrapping around this way and if I take the fingers of my right hand and wrap it in that direction around the wire, my thumb points to what is going to be the magnetic north of my electromagnet, which is another right-hand rule. 0778

By the way, if inside the solenoid, I place something like an iron bar inside it, it is going to get much, much, much stronger, so that electromagnet -- if you put something like iron inside it, gets much stronger -- a way to greatly increase your magnetic field strength. 0795

Now let us talk about forces on parallel wires. 0815

Current carrying parallel wires exert forces on each other. Why do they do that? 0817

Well, if you have two wires that are parallel to each other, the current through one wire is going to create a magnetic field and we just determined that. 0822

Now the other wire has moving charges and moving charges going through a magnetic field observe a force. 0830

They are going to create a force on each other. 0836

A typical sort of AP question might be something like: If we have a wire with current flowing that way and another wire right beside it with current flowing toward it, what are you going to see for force between these? Which direction? 0840

The way I like to think about it is let us start with this one on the left. 0853

If we have a current flowing this way -- If we think about the direction of the magnetic field -- do the right-hand rule -- I am going to have the magnetic field coming toward me on this side and as it comes around it is going to be going in to the plane over here. 0856

Now the current here, I could use another right-hand rule. 0869

Positive current -- point my fingers up in this direction, bend in the direction of the magnetic field due to the other one and I am going to see a force to the left on that wire. 0875

Now try the same thing in the opposite direction to prove that you are going to see a force on that original wire back toward the middle. 0885

If we have current flowing in the same direction, we get an attractive force. 0891

On the other hand, if we have two wires where the current is flowing in opposite directions, you are going to see that they have a repulsive force; they are going to force each other away. 0896

Try proving that as a really great practice making sure you understand how some of those right-hand rules work. 0910

Let us take a look at the magnetic field due to two wires. 0919

Two different wires carry current as shown below. 0922

Find the magnetic field here at point (P) in between them. 0925

Well, let us call this on the left, wire 1 and on the right let us call that wire 2. 0929

The way we are going to find the net magnetic field at (P) is let us find the magnetic field due to I1 first, then we will find the magnetic field due to I2, and we will add them together. 0937

The magnetic field due to 1 is going to be μ0I/2πr, which is 4π × 10-7 × 4 A (current)/2π and the distance from our wire to point (P) is .15 m.0947

When I do that, I am going to find that B1 is equal to about 5.33 × 10-6 T. 0968

For the direction, by the right-hand rule, I will point my thumb in the direction of positive current flow, wrap my fingers around and I am going to see that that is going to be into the page at that point. 0978

Now let us do the magnetic field strength due to wire 2. 0990

B2 is also μ0I/2πr or 4π × 10-7 × 3 A/2π and our radius now, our distance from the wire is .1 m. 0996

When I put that in my calculator, I find that B2 is about 6 × 10-6 T, and as far as direction goes, let us do the right-hand rule again. 1013

Point the thumb of my right hand in the direction of positive current flow, wrap around that imaginary wire and I see that that is going to be coming out of the plane of the page. 1024

That is going to be going in this direction; they are in opposite directions. 1032

If I want the total magnetic field strength at (P), that is going to be equal to the magnetic field strength due to 1 plus the magnetic field strength due to 2, which is going to be equal to...1038

...well B1 was 5.33 × 10-6 T and that is coming out of the page. 1053

We also have this 6 × 10-6 T or that is going in to the page, 6 × 10-6 from B2 is coming out of the page toward me. 1064

Well notice out of the page is stronger, so I am going to subtract them, -5.33 × 10-6 + 6 × 10-6, to give me a total of about 6.7 × 10-7 T...1076

...and because out of the page is stronger, the net magnetic field must be out of the page, so there is my answer -- the magnetic field at point (P) is 6.7 × 10-7 T out of the page. 1091

Let us take a look at a problem with a strength of an electromagnet. 1115

We have a solenoid of length 10 cm containing 150 coils of wire and carrying a current of 10 mA. 1119

Determine the magnitude of the magnetic field created inside the solenoid. 1126

Well the magnetic field strength due to a solenoid is μ0 × N × I, where (N) is the number of coils per unit length. 1130

That is going to be 4π × 10-7 μ0 × 150 coils/.1 m × 10 mA (current flow) = .01 A...1140

...so I would get a magnetic field strength of about 1.88 × 10-5 T. 1157

Let us do an example with a force on a wire. 1170

We have a 5 m long straight wire -- so L = 5 m -- which runs at a 45 degree angle -- θ = 45 degrees -- to a uniform magnetic field of 5 T, so B = 5 T. 1173

If the force on the wire is 1 N -- F = 1 N -- find the current (I) in the wire. 1188

All right, the force on the wire, FB, is equal to ILB sin(θ) and we are looking for current, therefore current is going to be the force over LB sin(θ)...1196

...which is going to be 1 N/5 m, our 5 T field and sin(45 degrees), which is going to be square root of 2/2. 1216

Therefore our current is about .0566 A or 56.6 mA. 1230

One more problem -- Determine the direction of the force on the two current carrying wire shown below. 1246

Well this comes straight from our previous slide where we were talking about what the force is between two current carrying parallel wires, whether the current is in the same direction or opposite directions. 1253

You can figure this out by working through the right-hand rules or remembering that in opposite directions you have a repulsive force between them. 1264

Hopefully that gets you a good start on magnetism as it relates to current carrying wires. 1278

Thanks so much for your time and for visiting us at Educator.com. Make it a great day everyone!1284