A rectangular loop of wire carries a 2.0-A dc current and lies...

Question

A rectangular loop of wire carries a 2.0-A dc current and lies in a plane which also contains a very long straight wire carrying a 10.0-A current as shown in Fig. 28–58. Determine (a) the net force and (b) the net torque on the loop due to the straight wire.

Accepted Answer

Everyone. Let's take a look at this practice problem dealing with magnetic forces. This problem says, consider two conductors given in the diagram below the length and width of the loop are six centimeters and four centimeters respectively. This loop carries a direct current of magnitude of five amps. The second conductor is an infinitely long wire with direct current of 15 amps. We need to compute the net force and the net torque exerted by the infinitely long wire over the rectangular loop. And below the question, we're getting a diagram that has a rectangular loop with a counterclockwise current going through it. And the bottom of that rectangular loop is located five centimeters above the infinite long wire which has a current going to the right. We also again four possible choices as our answers. For choice A we have F net is equal to 8.0 multiplied by 10 to the negative six newtons towards the wire. And tau net is equal to zero. For choice BF net is equal to 8.0 multiplied by 10 to the negative six newtons away from the wire. And Tau net is equal to zero for choice CF net is equal to 3.0 multiplied by 10 to the negative six newtons towards the wire. And tau net is equal to five multiplied by 10 to the negative six newton meters. Every choice DF net is equal to 3.0 multiplied by 10 to the negative six newtons away from the wire. And how net is equal to five multiplied by 10 to the negative six newton meters. Now, the first thing we want to do is figure out the direction of the magnetic field that's being generated from the infinite long wire. And then use that to find the direction of the forces acting on each segment of a wire of the loop. So we'll start off by looking at the magnetic field. And for that, we need to recall our right hand rule for finding the direction of the magnetic field due to an infinite long wire. And for that, I'm going to point my thumb in the direction of the current. And I wanna curl my fingers in the direction of the magnetic field recalling that they form circles around an A wire. When I do that, when I reach the plane that has the loop in it, my fingers are pointing towards me. And so that means that the magnetic field in the plane where the loop is located is pointing out of the screen will indicate that on our diagram. So we have the direction of the magnetic field we can use it to find the direction of the forces on each segment to recall the force of um acting on a segment due to the magnetic field F is going to be equal to I multiplying the quantity of L cross B where I here is the current L is going to be the length of the segment of wire and B is the magnetic field. And the direction of L in this case, the vector is in the same direction as the current is flowing. So we're just going to use this cross product to figure out the direction of the force. So I wanna start off with our bottom segment of the loop. And here the current is going to the right, the magnetic field is coming out of the screen. And so when I do the cross product, I point my fingers in the direction of my first vector, which is the L which is going to the right I curl my fingers in the direction magnetic field which is coming out of the screen and my thumb points in the direction of the force which in this case is pointing downwards. So we'll label that on our diagram and we'll label that force F one, we'll do the same thing for the other segments. So on the right uh hand segment of the loop, when I do the right hand rule, I'll find that the force is pointing to the right, we'll call that F two for the top segment of the loop to do the right hand rule, the force is pointing upwards, we'll call that F three. And for the left hand segment of the loop, when I do the right hand rule, that force is pointing to the left, we'll call that F four. So now that we have the directions, in order to find the net force, we just need to find the magnitude of each of these forces and then add them together. So in order to find the magnitude of the net force, we need to find the magnitude of the magnetic force for each segment. And for that, we'll need the magnitude of the magnetic field that's being generated by the infinite long wire. So recall your formula for the magnetic field that will be B is equal to, you know, I divided by two pi R where I here is the current that's flowing through the wire. And R here is the distance from the wire to whatever point I'm looking at and B is the magnetic field. So we're gonna use that to calculate the magnitude of the force. So I'm gonna rewrite my force equation as a magnitude. So it'll be F is equal to. And here I have to be careful about the currents, but I'm gonna have a current in both the loop and the log wire, we'll call the current here in the loop il or I loop. And this gets multiplied by the length of the segment that we're gonna be looking at label as L multiplied by the magnetic field B. And this gets multiplied by the sine of 90 degrees, which is the angle between the L vector and the magnetic field vector B for all four segments, it's always perpendicular. So we have the sign of 90 degrees since the time of nine degrees um is just one, we can rewrite this as F is equal to il multiplied by L. And here, I'm gonna plug in for my magnetic field and that's gonna be the mu knot multiplied by I wire, which will label as IW divided by two pi R. So this is gonna be the magnitude for each of the segments of loop. And what I need to do here now is I'm gonna start by looking at the magnitudes of forces F two and F four. So if I look at our equation here for both F two and F four, they're gonna have the same length, they're going to have the same currents running uh through the loop, it'll have the same current through the wire. And they're also the same distance from the wire from the infinitely long wire. So even though R is going to vary, it's going to vary in the same fashion across both of those lengths. So that means that these two forces are gonna have the same magnitude. But if I look at the diagram, I have them going in opposite directions. F two is going to the right F four is going to the left. So that means that they're going to cancel out. So I don't need to worry about calculating what those, the values of those forces actually are. They're going to cancel out. So really, my net force is only going to depend upon F one and F three. That's what we're going to focus on. So if I come down here and look at my net force, we'll have F net is equal to F one plus F three. And here I need to define a positive direction. So we just have the positive direction going upwards. So I'm gonna plug in our formula or are two different forces. So I'll have F net and since I'm working in just one direction, I'm gonna drop the vector notation. So this is gonna be equal to and for both F one and F three, I wanna have the same constant. So I'll have the mu knot divided by two. Pi I'm gonna have the same current, I have the il and we'll also have the same current in the wire. So we'll have that multiplied by IW and they're gonna have the same length. So they have the same L. So the only thing that's different here is actually the distance from the wire uh from the infinite long wire to the wire segment in the loop. So I'm gonna factor out the Muno divided by two pi and that's gonna be multiplied by il multiplied by IW, multiplied by L. And then I'll have that multiplying the quantity. And for F one that's pointing downward, so it's gonna go in the negative direction, I'll have one divided by R one. So it's distance from the infinitely long wire, then I'll have plus one divided by R three. So at this point I can plug in values and I'll have F net is equal to. And for Muno, we have or pi multiplied by 10 to the negative seven that has units of tesla meters per amp that gets divided by two pi and this gets multiplied by the current in the loop, which was five amps. This is multiplied by the current in the wire which is 15 amps. This gets multiplied by the length of the segment. And for both um F one and F three, the length of that segment of the rectangular loop was six centimeters. And I'm not going to convert that from centimeters to meters. Um Because my uh next quantity I'm gonna be putting into the formula is also going to have units of centimeters. And so it'll cancel out. So any conversion factor that would do will also cancel out. So this quantity is now multiplied by the quantity of minus one divided by. And for R one, if I look at my diagram, that's gonna be the distance from the wire to the lower end of the segment, which was five centimeters, then I'll have plus one divided by. And here it's going to be the distance from the wire to the top segment of the loop. And that was the five centimeters to get to the bottom loop plus the four centimeters separating the top and bottom of the loop. Down on the right hand side, I just have a numerical value. And so I want to plug that into my calculator. I get F net is equal to minus 8.0 multiplied by 10 to the negative six newtons. Here we have to interpret what that negative sign actually represents. And since the positive direction was going up, that means that the net force is pointing downwards, which means it's pointing towards the infinitely long wire. That's gonna be the first part to my question answer to my question. Now, for the second part, I need to recall our definition for torque. So we call that Twerk tao is equal to RF sign data. So here I'm looking at just the magnitude of the torque. And so R here is going to be the distance from the axis rotation to whatever the force is F is the force and theta is the angle between the R and F vectors. Now, if I look at this loop, my forces are all in the same plane as my loop and they're exactly opposite of each other. So what's gonna happen here is if I look at say the torque about the center of my loop, that means that the angle theta between the R and the F for all four of my forces is going to be zero degrees. And so the torque due to each of the forces is gonna be zero. So that means, but my net torque is also going to be zero. So since my net torque is going to be zero, for that one, it's actually gonna be the same for all the other cases, it gets a little bit more mathematically involved because the force is not being um active to one at a single point on each of the segments, it's actually being exerted over the total segment. So the calculation becomes a little bit more involved, but you end up with the same result that when you add up all the torques, it actually equals zero. So that means that the net torque on this loop is going to be zero. Another way to think about it physically is each of the forces are just trying to stretch that loop. It's not trying to spin it in any way. And since it's a metal loop being made of conducting material, it's not actually going to bend or stretch. And so therefore, it's not going to rotate. So when I take both of those answers together, that the net torque is equal to zero and that the net um forces the 8.0 multiplied by 10 negative six newtons pointing towards the wire. And I look at my answer choices, this corresponds to answer choice. A so just a quick little recap of what we've done here. We used the right hand rules to figure out the directions of our magnetic fields and our forces acting on the loop. And then we use our definition for the magnetic force on a wire as well as the magnetic field to do to an infinite long wire to come up with an expression for the magnitude of each of the forces acting on this, each segment. We then um added those together and we found our net torque. And we also looked at uh sorry that we found the net force. We also looked at the net torque and we found that that was equal to zero. That's basically because our forces are trying to stretch the loop instead of trying to rotate it. So I hope that this has been useful. And I'll see you in the next video.

A rectangular loop of wire carries a 2.0-A dc current and lies in a plane which also contains a very long straight wire carrying a 10.0-A current as shown in Fig. 28–58. Determine (a) the net force and (b) the net torque on the loop due to the straight wire.

Key Concepts

Magnetic Field Due to a Current-Carrying Wire

Lorentz Force

Torque on a Current Loop

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