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

### 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
- Objectives
- Force on a Current-Carrying Wire
- 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
- Example 2
- Magnetic Field Due to a Current-Carrying Wire
- 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
- Force Between Parallel Current Carrying Wires
- 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
- Example 4
- Example 5
- Example 6
- Example 7
- Example 8
- Example 9

- 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

### 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: 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 indicating *0098

*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 torque *0151

*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 and *0278

*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 create *0298

*magnetic fields but they also interactive magnetic field so you can insert forces upon each other *0303

*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 or*0380

*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 wire *0447

*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 wire *0511

*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 flux *0578

*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 field *0638

*due to these wires at point P located halfway between the 2 wires that the top wire carries a current of 3 amps *0645

*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 force *0778

*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 out *0806

*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 force *0819

*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 × 10⁻⁷ *0876

*× 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 tesla *1295

*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 bar *1382

*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 and *1422

*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

1 answer

Last reply by: Professor Dan Fullerton

Mon Jan 18, 2016 7:17 AM

Post by Parth Shorey on January 17 at 04:06:59 PM

Why are we always using the right hand rule, and not the left hand rule?