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Magnets and Magnetic Fields | 22 mins | 0 completed | Learn |

Summary of Magnetism Problems | 10 mins | 0 completed | Learn |

Force on Moving Charges & Right Hand Rule | 31 mins | 0 completed | Learn Summary |

Circular Motion of Charges in Magnetic Fields | 18 mins | 0 completed | Learn Summary |

Mass Spectrometer | 34 mins | 0 completed | Learn Summary |

Magnetic Force on Current-Carrying Wire | 23 mins | 0 completed | Learn Summary |

Force and Torque on Current Loops | 18 mins | 0 completed | Learn |

Concept #1: Force and Torque on Current Loops

**Transcript**

Hey guys. In this video we're going to talk about magnetic forces and magnetic torques when you have current loops, let's check it out. Remember that if you have a current carrying wire in other words a wire that has current running through it and we have that wire in a magnetic field it will feel a magnetic force. So a current in an existing in the presence of an existing field feels a force and that force was given by Bil sine of theta. Okay. This is old news. Now realize that wires don't have to be straight obviously, in fact in circuits they turn around all the time but you can actually get a wire to also look like a loop. So imagine a long wire that you just bend so that it looks like this. So you maybe have some current going this way which means current going that way, this way, all the way around and current leaves this way. And who knows, maybe this is connected to a battery somewhere, right, so that it produces a current. Anyway we have a wire there with a current and in some cases when you have this arrangement you're going to get a torque. Not always but in some cases. First things first. The net force is always going to be 0 so you should remember that whatever forces exists in this wire will cancel each other out and then we're gonna get a magnetic torque. But first let me show you how these forces are going to work. So that the torque can make sense. So this is a wire so we're gonna use the right hand rule to look at directions of forces. This is the magnitude of the force right here but we're gonna look at the direction of force. So I'm gonna follow this path here and I have my right hand that I wanna go right, okay. And the area that we're actually interested is sort of like this square area. So let's star over here, let me delete that green right there. So we're gonna start with the first arrow. I'm going up, right, and the magnetic field at this point is also up so the only way I can do that is if I do this and then I put this up, right, or this, right. And this is bad news, if you remember right hand rule you're supposed to be like this, this is ideal maximum force. You can do this right, that's cool to, that's a little bit less force and this gives you a force of zero and that's okay, that's actually what's going on here. And this wire segment over here, let's call this current in part one. All the current will be the same but this is just part one. I'm gonna say, actually this is called left. So I'm gonna say that the force on the left segment of this wire will be 0. Because the current is parallel with the magnetic field. Magnetic field the blue lines here, okay. And that's cool. But now we're gonna go this way which means my thumb is gonna go to the right and if I want my B fields to go up, look what happens. My hand, please do this and follow me. My hand is, look at my hand my palm is going this way which means it's going out of the page so the force on this wire will be out of the page and out of the page is represented by little dots. Okay. So the force on the top segment of the wire will be out of the page. When you go down, what you're gonna have is something similar on the right side and you had in left. So I'm going down but then my B lines are supposed to go up and the only way to do that is to do this. I make a 180 degree there and we know there's no force in that situation. Either force on the right side of the segment of this wire will be 0 and then at the bottom here I'm going to the left and I want my B field to be going up which means my hand is away from me and into the page into the page into the page looks like this, okay. So the force on the bottom is into the page. So look what happens. Look at this rectangle or this square, whatever. This top part is being pushed out while the bottom part is being pushed in. So you than do is imagine that you can fix this around a axes of rotation, right, you can fix the square and axes of rotation. The top part of the square is being pushed out which means out of the page means towards me when the bottom is getting pushed into the page which means away from me. The top comes in and the bottom comes out so this thing does this, right. Which means it's going to rotate. That's why I get torque. Remember torque is a force away from a axes od rotation in such a way that you have a rotation so that's what this thing is gonna do. This thing is gonna start spinning and you have a torque. That's how it works and if you do some calculation you would write the equation which i'm just going to give to you and the equation looks different from what I'm gonna write but I write it this way so that it's easier to remember which is N B A I sine of theta N B A I sine of theta so I have to remember there is an I sine of theta there, don't forget that part. The torque equals N B A. N is the number of loops, we'll talk about that in a second just write that there. B is the strength of the magnetic field that's a factor. A is the area, is the area that is made by the wire arrangement, right. So this is the area, I is the current and sine of theta. Theta is the angle between the A and the B. Okay, so theta the angle of theta is between the normal of the A and B. We'll talk about the normal of the area in a second. Number of loops. What's the deal with the number of loops. Well you can get a wire, you can get a wire and make a little loop, something like that, right, or you can get a wire and make 2 loops on top of each other, it would look exactly like this because it's sort of like the first floor and the second floor of loops, the only difference is that here N equals 1 and than here is N equals 2. A lot of the times you first learn this equation without the N and than they say hey you can do multiple loops and than they add the N, I'm just putting the N already in there and that way you get the N yey you don't have to go back in this equation. Cool, so that's that. By the way most the times N is going to be 1 because you're gonna have a single loop in fact they don't specify the multiple loops. N will be 1 so 1 unless otherwise stated. The normal of A. Remember whenever you have an area, whenever you have an area so imagine this like a surface of plane or area, right, the normal to the area is normal means perpendicular which means 90 degrees is 90 degrees away from this surface. So this is your area the normal of A is this. Right, I'm gonna try this in a bunch of different angles. If you get something like a textbook and you have something come out of it, imagine if I poke a hole in this which I'm not gonna do, something like that. Okay so that's the area. Always 90 degrees to the surface. Hopefully that helps. The normal of A and the direction of B. We're gonna talk about this a little bit more but this is the big equation that really matters here. Lastly there's this property this quantity called magnetic moment, I'm not gonna get into it but I just want you the magnetic moment given by the letter mew, not to be confused by the constant mew not. Right, given by the letter mew, which the same letter for coefficient of friction but totally unrelated, is the equation for this guy is N A I, okay. So this is N B A I and this is nine or name or something like that. So mew is N A I and if you look at this equation this has an N, this has an A and this has an I so technically you can rewrite torque as mew because it's gonna take over for these 3 guys, B sine of theta. It's another equation for torque. You really have to remember that equation because if you know these 2 you should be able to manipulate and come up with that 3rd one, okay. So that's the magic moment the magnetic moment rather. Anyway this is simpler than it looks, it looks hairy but it isn't, let's do an example here. So it says here a loop which looks like this with magnetic moment, oh that's that guy mew equals 0.5 carries a current of .01 so I equals .01 it is placed in the presence of a magnetic field of strength 0.05. 0.05 or just 05 and then which points in the plane of the loop, we wanna know what is the magnitude of the torque, okay. And than this is the hard part. Here is this which points in the plane of the loop so it didn't turn exact direction of the loop but let's just draw a loop. By the way the loop doesn't have to be rectangular it could actually look like something like this and it works just the same. It's saying here that the magnetic field points in the plane of the loop. The plane of the loop is this, right. Is this so it means that the magnetic field looks something like this or something like this or something like this. We don't know, it doesn't say so we're just gonna pick one. I'm actually just draw them up so that it looks more similar to what we already have here. In the plane of the loop so back to my little surface here. The plane of the loop is this, right. If it's horizontal it's the floor or if it's like this the wall so it means that the magnetic field is pointing in this direction here. Okay. In this direction and remember A is perpendicular to that. So what's the angle between the plane of, the plane of the loop and the normal the loop it's always 90 degrees. Okay. So let's break it down. The angle between the plane and it's normal is always 90 degrees. Because it's if you have the plane down here, this is the direction of the plane and this is the direction of the normal. Okay so this is my B, my A is out of the plane so my A is either into the page or out of the page, we don't know so I'm just gonna say that it's out of the page, right. What matters that the angle between A, B and the normal of A is 90. Okay so the angle I'm gonna use in the equation is going to be the sine of 90 which by the way is one. Hopefully it makes sense. I spent a lot of time on the angle there because that's the hard part, everything else is easy. How do I find torque. I might be tempted into writing the big equation N B A I sine of theta and I have theta, by the way this whole thing is 1 so let's get rid of it. I have, it didn't say anything about loops, it just said a loop, a loop so N is just 1 because otherwise, B I have it, A I don't have it, I I have it. We're stuck, I don't have A, we can go find A or because we have mew, we can just plug it into this equation. So this was a little tricky but it's just super basic physics problem solving. Good hustle here, mew B sine of theta just 1, mew is .5, B is .05, this is 0.025. the union for torque you may remember from a long time ago torque is Fr, that's like the general equation for torque. So the units are Newton times R is a distance Newton times meter. That's it for this one, let's keep going.

Example #1: Torque on a Loop at an Angle

**Transcript**

Hey guys so let's check out this example. So here I have a wire arranged in a loop and that's this blue line here, this is the loop the current loop and we wanna know what must be the current running through that loop given all this information. So I wanna look for the current. Cool. So it says here that B is arranged as a rectangular 4 meter wide and 2 meter deep. I forgot to say that this is a loop but these 2 pieces of information 4 meters and 2 meters will allow you to find the area. So right away we know that the area is 8 square meters. It is placed in the plane shown where constant 5 T magnetic field exists. B equals 5. It says the wire loop is parallel to the plane, parallel means side by side. So imagine and x acts right here and I'm gonna do this just to try to sort of try to support that image. Imagine the x acts as your parallel so you're sitting on a plane. If it's a horizontal plane which it is here, looks like it is, than you're essentially on the floor so the cable is on the floor. Imagine that this is the floor the x is plane and this is sort of the height here. So it's parallel to the plane and the magnetic field is directed 30 degrees above the plane. So if this is your plane than the magnetic field are lines that go like this so instead of them going like this they're going like that, okay. So it's a little weird to draw here but basically were gonna have magnetic field lines that look like this and they make an angle of 30 degrees this is like the positive x axes. It says here they make an angle 30 degrees above the plane, okay. Above the plane. The loop experiences a torque of 10 and we wanna know what is the current. So how do we do this. There's 2 equations for torque, you can use N B A I sine of theta or you can use the magnetic moment B sine of theta. Okay. Here we don't have the magnetic moment, we can calculate it but there's no point in doing that, it's easier just to plug everything here. I'm looking for I, I equals torque divided by N B A sine of theta. I have torque which is 10, N is the number of loops, it didn't specify how many loops we have so we're gonna assume that it's 1 that's what we're supposed to do. B is 5 area is 8 and as usual the tricky part is finding the angle. So which angle should we use. Remember the definition is that it's the angle between A and B. Now whenever we talk about the direction of an area it's always the normal of that area meaning if you have an area it's the direction that is perpendicular to that area the direction that makes an angle of 90 degrees with it. Okay, so it's this, okay. This area. So if this is the area here, and the area enclosed by the wires, that is sitting on an x, on the floor, the perpendicular to that will be going up like this. Okay. Going up like this in other words the A vector points in this direction here. This is the normal of the surface, okay. Normal of the surface. The angle I want is between the normal of the surface and B and this is B right here. Okay, so they sort of cross right here so what I want is I want this angle here. This definition. By the way I noticed I almost always give you the wrong angle because I want you to be very careful I want you to be paranoid about this stuff but just know that it's not the case that it's, that you're always gonna get the wrong angle. You could very well just get lucky and get the right angle or maybe some sort of like backwards tricky question that you actually get the right angle and maybe you change it. So the angle here is 60 because that's the one I want and we're gonna put it here, 60. So if you do all of this and I have it here, the answer is gonna be .29 or just .30 or .3 amps. Cool. That's it for this one, do we wanna know the direction of the current. We're not told the direction the torque, we can't find that. Never mind. So that's it for this one, let's keep going.

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Concept #1: Force and Torque on Current Loops

Example #1: Torque on a Loop at an Angle

A rigid, rectangular loop, which measures 0.30 m x 0.40 m, carries a current of 7.7 A, as shown. A uniform external magnetic field of magnitude 2.6 T in the negative x-direction is present. Segment CD is in the xz-plane and forms a 33° angle with the z-axis, as shown. The magnetic moment of the loop is closest to?
A) 0.31 Am2
B) 0.77 Am2
C) 0.46 Am2
D) 0.62 Am2
E) 0.92 Am2

A circular coil of wire 10.0 cm in diameter has 20 turns and carries a current of I = 10.0 A. The coil is in a region where the magnetic field is B = 0.25 T. What orientation of the coil gives the maximum torque on the coil?

A solenoid with N = 16 turns, length L = 0.30 m, and cross-sectional area A = 2.0 x 10 -3 m2 sits at an angle Φ = 30° with respect to a uniform magnetic field of magnitude B = 2.0 x 10-4 T, as shown in the figure. The solenoid carries a current I and experiences a torque of magnitude 6.4 x 10-9 N•m. What is the magnitude of I?
(a) 1.2 mA
(b) 0.6 mA
(c) 32 mA
(d) 2.0 mA
(e) 0.5 A
(f) none of the above

A square wire 2.0 cm on each side is oriented so that a 5.00 mT magnetic field makes a 30.0° angle with the normal to the plane of the square. If a 1.0 mA current causes a 30.0 Nm torque on the wire, how many loops does the wire make?
A) 4.2 x 1010
B) 1.5 x 1010
C) 3.0 x 106
D) 3.0 x 1010

The coil in the figure below lies in the y-z plane, has a current, I, in the direction shown and is free to rotate about the z-axis. What should the direction of an external B-field be to cause a clock-wise rotation as shown?
A. Positive y-direction
B. Negative y-direction
C. Positive z-direction
D. Negative z-direction
E. Negative x-direction

A copper wire is 8.0 m long and has a cross-sectional area of 1 x 10 -4 m2. The wire forms a loop in the shape of a square and is connected to a battery that applies a potential difference of 10 V such that current flows clockwise (looking down along z-axis) as shown below. The loop is placed on the XY plane in a uniform magnetic field of magnitude 0.5 T directed along the X direction (battery not shown). What is the magnitude of the torque on the loop and its expected direction of rotation? Resistivity of copper is 1.7 x 10-8 Ωm.
A. 14706 N.m, rotate about y axis
B. 14706 N.m, rotate about z axis
C. 24802 N.m rotate about y axis
D. 24802 N.m rotate about z axis
E. None of the above

An electric current runs through a coil of wire as shown. A permanent magnet is located to the right of the coil. The magnet is free to rotate. What will happen to the magnet if its original orientation is as shown in the figure, with the current coming in on the front side of the solenoid, and then looping around the back?
1. rotate clockwise
2. remain still
3. rotate counterclockwise
4. Unable to determine

A rectangular loop with L1 = 0.15 m and L2 = 0.35 m is sitting in a magnetic field B = 0.35 T as shown in the figure. There is a counterclockwise current I = 0.55 A in the loop. Part (a) Express the magnitude of the magnetic force on ab or cd, Fl, in terms of the length Ll, current I and magnetic field B.Fl =Part (b) Calculate the numerical value of the magnitude of the force, Fl, on ab or cd and N.Part (c) Express the magnitude of the magnetic force on bc or ad, F2, in terms of the length L2, current I and magnetic field B.

A rectangular loop with L1 = 0.15 m and L2 = 0.35 m is sitting in a magnetic field B = 0.35 T as shown in the figure. There is a counterclockwise current I = 0.55 A in the loop. Part (a) Express the torque of Fl ab, with respect of the axis ef, in terms of Fl and L2. Part (b) Express the torque of Fl on cd, with respect of the axis ef, in terms of Fl and L2.Part (c) What is the total torque on the current loop with respect of axis ef, in terms of F, and L2.

Consider the wire loop you used in part A. For the lab, you placed the bottom section of the loop in the magnetic field (between two magnets in a box with opening in between for loop to go in) and measured the force upon it when current was flowing through the wire. Imagine instead that both the bottom and one side of the loop were placed in the magnetic field (but not the top or other side). What would be the net direction of the force upon this loop? Explain carefully and refer back to equations if needed. Please specify the NET direction!

The 20.0 cm by 35.0 cm rectangular circuit shown in the figure is hinged along side ab. It carries a clockwise I = 5.35 A current and is located in a uniform 1.45 T magnetic field oriented perpendicular to two of its sides, as shown.a) Which force exerts a torque about the hinge ab?b) Calculate only the force that exert the torque mentioned in part A.c) Use your results from part B to calculate the torque that the magnetic field exerts on the circuit about the hinge axis ab.

A uniform rectangular coil of total mass 270 g and dimensions 0.500m×1.00m is oriented parallel to a uniform 3.70-T magnetic field (Figure 1) . A current of 2.00 A is suddenly started in the coil.Find the initial angular acceleration of the coil just after the current is started.Express your answer using two significant figures.ANSWER = _________________ rad /S2

Consider a 200 turn square loop 22 cm on a side.Randomized Variablesl = 22 cmCalculate the magnetic field strength (T) needed on the loop to create a maximum torque of 275 Nm if the loop is carrying 21 A

A square current loop 5.2 cm on each side carries a 480 mA current. The loop is in a 0.70 T uniform magnetic field. The axis of the loop, perpendicular to the plane of the loop, is 30° away from the field direction.What is the magnitude of the torque on the current loop?

People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means that unrealistically large currents are needed to produce noticeable torques. Suppose a 20-cm-diameter loop of wire is oriented for maximum torque in the earth's field.What current would it need to carry in order to experience a very modest 1.0 × 10−3 N⋅m torque?

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