Subjects

Sections | |||
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Magnetic Field Produced by Moving Charges | 11 mins | 0 completed | Learn |

Magnetic Field Produced by Straight Currents | 29 mins | 0 completed | Learn Summary |

Magnetic Force Between Parallel Currents | 13 mins | 0 completed | Learn |

Magnetic Force Between Two Moving Charges | 9 mins | 0 completed | Learn |

Magnetic Field Produced by Loops and Solenoids | 43 mins | 0 completed | Learn Summary |

Toroidal Solenoids aka Toroids | 12 mins | 0 completed | Learn |

Biot-Savart Law with Calculus | 16 mins | 0 completed | Learn |

Ampere's Law with Calculus | 17 mins | 0 completed | Learn |

Additional Practice |
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Ampere's Law |

Concept #1: Magnetic Field Produced by Moving Charges

**Transcript**

Hey guys so in this video we're gonna talk about moving charges producing a magnetic field. Let's check it out. So remember that a charge if a charge moves through an existing magnetic field it's going to feel a magnetic force. And we've seen this before, you have magnetic field B right here and you have a charge q that moves through it with a speed v, this charge will feel a force, this is called the Lorentz force and you may even remember the equation is q v B sine of theta. Super important equation, got to know it. What I also need you to know, and remember, is that not only will the charge feel a force if it goes through the existing magnetic field but a moving charge is also going to create it's own magnetic field away from itself. Okay so moving charge is also going to produce, produce a new magnetic field. Okay. You need to know that these 2 thing happen, you need to know that they happen at the same time, it's not one or the other it's both. Now I need you to know that but I also need to have a disclaimer here, big disclaimer that this topic is much less popular than for example being asked to find a force that a moving charge feels. Okay. In fact most professors and most textbooks skip this all together and the reason IÕm including this in the video across all the textbooks we cover, even a lot of the textbooks don't have this topic is because it's a really big point and I think it's very helpful for you to remember that a moving charge can feel a force and produce a magnetic field. That you have to know no matter what but you may not need to know is what follows is how to actually use this equation to solve a problem for this specific kind of question. Okay, so here's the equation I give you and what you need to do is you need to get this equation and go talk to your professor if you're not sure if you know this and ask hey do I need to know this and if they say no than problem solved or if they look really confused or they don't know where you got this equation from what are you talking about than now you know that you don't have to know this, okay. So the equation it's another one but it's pretty straight forward it's relatively straight forward to plug stuff in, B equals mew not, mew not is a constant right here, times q v sine of theta divided by 4 pi r squared. Mew not is a constant, q is charge, v is speed, sine of theta, I'll talk about theta over here, 4 pi r squared, r is the distance, okay. Now these two points here talk about how to properly use this point here talks about how to properly how to figure out theta to properly use this equation and this one talks about how to find the direction of the magnetic field. I'm just gonna jump straight into this example because rather than talk about these it's much better to show you how they work in action, okay. So I have a 3 Colon charge let's draw a little q equals 3 Colons and it's moving to the right with the speed v equals 4 meters per second and we wanna know the magnitude and the direction of the magnetic field so what is B and what is it's direction. Our charge produces 2 centimeters directly above itself so here is a charge I wanna know what is the magnetic field if this is point p at a distance 2 centimeters our distances is r so r equals 0.02 meters I wanna know what is the magnetic field at this point p produced by this charge here. And obviously the equation we're gonna use is this B equation right here and I'm gonna write that B equals mew not which by the way if you want you can already replace it with this so let's do that. Mew not is gonna be 4 pi times 10 to the negative 7 you always want to write the left version not the right version because the left version you're going to cancel out with 4 pi in the bottom of that equation right away and then I have q v sine of theta. Q is 3, v is 4, and sine of theta we'll talk about theta in just a second. Divided by four pi I already canceled that times r square so 0.02 you can also write this as 2 times 10 to the negative 2 squared I think it looks easier to manipulate the numbers. So the only question here that's kind of tricky is what is your theta. Okay. What is your theta. And it says here theta is the angle between the v vector and the r vector and the r vector is a vector between the charge and the location of the produced field. You can think of the location the produced field as the target, target location. So if r is the vector between the charge, the charge is right here, vector just means an arrow and the location of the produced field. I wanna know what is the magnetic field here. That is my target location. I wanna know what is the magnetic field there so I'm gonna draw a line from the charge, it says right here, from the charge to the target location. So I'm gonna draw this line here, this is my r vector. And this r vector is only useful so that I can figure out what is the angle between the blue arrow and the red arrow and this angle is of course 90 degrees. So this is gonna be sine of 90, a whole lot of work for nothing because sine of 90 is just 1 but obviously you had to figure that out. And now we can simplify some stuff here. So I'm gonna leave 3 and 4 by themselves times 10 to negative 7 and then this 2 is gonna be squared which becomes a 4 and than the 10 to the negative 2 squared 10 to the negative 4. The 4 is canceled and I end up with 3 times 10 to the, gonna combine the exponents so it's negative 7 minus negative 4 so that's negative 7 plus 4 which is negative 3 Tesla because we're talking about magnetic field. Okay. So that's the field strength, the field magnitude. What about the direction. Now to solve the direction it says right here we want to use the right hand rule and by the way we would have used the left hand rule if q was negative so if q is negative which in this case it isn't. And what we wanna do is we wanna grab the line of motion. We wanna grab the line of motion so the best thing for me to do here is just show you how this works so let's move over here and we're gonna do that. So I wanna grab the line of motion so q is moving this way to the right and the line of motion is just a line formed by the direction of v so we can do this you can think the line motion like this, right, and what this does this line of motion separates the page into a top part and a bottom part. Okay. By the way if the v was moving up what it would do is it would separate this line, it would separate the page into left and right. That's the line of motion. Okay, so we got a line of motion there and we wanna grab it, we wanna grab the line of motion in such a way, since we're talking about the right hand rule, right, in such a way that my thumb points to the right, why. Because if you remember in the right hand rule your thumb is your velocity direction. Direction of your velocity so I'm gonna grab this this way and imagine that I can lift this here so that I can grab it and the only way to grab this, I can grab it like this, right, I'm gonna wrap my fingers around this line of motion I can grab it like this or I can grab it like this and we're gonna choose to grab it like this because we wanna point we want our thumb which is out velocity to point to the right so that it's consistent with the problem. When I do this the way to grab it is to go under, this is super important is to go under the marker over here and then come back up here so in the bottom half of this page I'm going into the plane to grab under the marker and then the top half I'm coming out of the page and towards myself towards my face and I have to do that, it's the only way I can grab in the direction I'm supposed to grab. So that means is that everything here is going to be into the page and everything here is going to be out of the page popping towards my face. You can put a bunch of little dots everywhere. Anything here, right, now we point somewhere over here which means, which means that the magnetic field at point p is going to be out of the page, okay. Out of the page. And that is the answer for direction. What if This was point p 1 and I want to know the magnetic field for point p 2 and maybe it's over here, doesn't even matter. The magnetic field for point p 2 here would be into the page. Into the page because it's on the bottom because when I grab this thing my hand goes in an than back out. Okay. The only difference from this is the last point, the only difference between into the page is the x, right. So the only difference between for example this point and this point is the magnitude because they're different distances and different angles but all of these points are in the direction that is into the page. Cool. Hopefully this makes sense, this is gonna come back later again when we talk about wires, when we talk about current flowing to wires, we're gonna grab the wire just like how I grabbed this line of motion. Alright, that's it for this one, let's keep going.

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Concept #1: Magnetic Field Produced by Moving Charges

A 6 μC charge moves with a velocity of. v = (1.2 km/s) î - (2.4 km/s) ĵ. What is the magnetic field, in vector form, at the point P indicated in the following figure?

A 15 C charge moves at a speed of 150 m/s in the –x-direction. What would be the magnitude and direction of the magnetic field at position P1 in the following figure? What about at position P2 in the figure?

A charged particle of mass m = 5.6×10-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 3.5 T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.83 m, 0) and leaves the region at (x,y) = 0, 0.83 m a time t = 776 μs after it entered the region.1) With what speed v did the particle enter the region containing the magnetic field? (m/s)2) What is Fx, the x-component of the force on the particle at a time t1 = 258.7 μs after it entered the region containing the magnetic field? (N)3) What is Fy, the y-component of the force on the particle at a time t1 = 258.7 μs after it entered the region containing the magnetic field? (N)4) What is q, the charge of the particle? Be sure to include the correct sign. (μC)5) If the velocity of the incident charged particle were doubled, how would B have to change (keeping all other parameters constant) to keep the trajectory of the particle the same?A. Increase B by a factor of 2B. Increase B by less than a factor of 2C. Decrease B by less than a factor of 2D. Decrease B by a factor of 2E. There is no change that can be made to B to keep the trajectory the same.

Part A.)Determine the direction of the force on the charge due to the magnetic field. (Figure 1). a) points into the page.b) points out of the page.c) points neither into nor out of the page and .d) .Part B.)Determine the direction of the force on the charge due to the magnetic field. (Figure 2)a) points out of the page.b) points into the page.c) points neither into nor out of the page and .d) .Part C.)Determine the direction of the force on the charge due to the magnetic field. Note that the charge is negative. (Figure 3)a) points out of the page.b) points into the page.c) points neither into nor out of the page and .d) .

An electron and a proton are each moving at 795 km/s in perpendicular paths as shown in (Figure 1) .Part AFind the magnitude of the magnetic field the electron produces at the location of the proton.Part BFind the direction of the magnetic field the electron produces at the location of the proton.

An electron and a proton are each moving at 795 km/s in perpendicular paths as shown in (Figure 1) .Part AAt the instant when they are at the positions shown, find the magnitude of the total magnetic field they produce at the origin.Part BFind the direction of the total magnetic field they produce at the origin.

A cosmic ray proton moving toward the Earth at 6.5 × 107 m/s experiences a magnetic force of 1.55 × 10-16 N .a) What is the strength of the magnetic field if there is a 45º angle between it and the proton’s velocity?

A proton moves along the x-axis with vx=1.0 × 107 m/s.a) As it passes the origin, what are the strength and direction of the magnetic field at the (0 cm, 1 cm, 0 cm) position? Express your answer in terms of the unit vectors î, ĵ, and k̂.b) As it passes the origin, what are the strength and direction of the magnetic field at the (0 cm, -2 cm, 0 cm) position? Express your answer in terms of the unit vectors î, ĵ, and k̂.

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