A long wire carrying 4.50 A of current makes two 90° bends, as shown in Fig. E27.35
. The bent part of the wire passes through a uniform 0.240-T magnetic field directed as shown in the figure and confined to a limited region of space. Find the magnitude and direction of the force that the magnetic field exerts on the wire.

Question

A long wire carrying 4.50 A of current makes two 90° bends, as shown in Fig. E27.35

Diagram showing a wire with 90° bends in a magnetic field, illustrating magnetic force on the wire.

. The bent part of the wire passes through a uniform 0.240-T magnetic field directed as shown in the figure and confined to a limited region of space. Find the magnitude and direction of the force that the magnetic field exerts on the wire.

Accepted Answer

Hey, everyone. So this problem is working with magnetic forces, let's see what they are giving us and what they're asking of us. So we have a segment of a current carrying conductor that is placed within a uniform magnetic field of a given magnitude. And that field is pointing directly out of the plane of the page. The conductor makes two right angle turns and is positioned as shown below here. And we are given the current and it flows within the conductor. So it is asking what the magnitude and direction of the magnetic force acting on the conductor is due to the magnetic field. So the first thing we can do is recall that the magnitude of the force is given by I L cross B when we're dealing with a uniform magnetic field that is working with a current carrying wire. So when we look at this equation, we can break out this problem into three different segments. So I'm just going to denote here that we're going to go from point A to point B and then down here is going to be C and then at the end is deep. And so the total force is going to be the sum of each of these segments. So from A to B plus the force from B to C plus the force from C to D. So when we write out the equations for each of these, We have our current, which is given to us as five amps. And then our length is going to be unique for each of these. So we have the length from A to B and that's gonna be in the I direction because we're going here and the positive X direction from A to B and then run across that with our magnetic field, which was given to us as 0.34 tesla and that's directly out of the page. So that's the positive Z direction. We're gonna write out the equations for each of these three segments and then talk about this length of term. So for B two C, we have the length that is working in the negative J direction, make sure have the negative sign there. And then J and then the same, I got it field and then from C to D, same current. And we're moving again in the positive X direction. So we have L from C to D in the positive I across the magnetic field. So when we add these together, we can simplify these length terms. So we were given from the problem, this length from B to see Is 40 cm. Remember you write that as a .4 m. And again, we've already said that in the negative direction, but we don't have from A to B or from C to D individually, but we do have them together. So L A D plus L C sorry, L A B bus, L C D Equals 80 cm or .8 m in the positive x direction. So now we actually have everything we need to solve for the magnitude of the force some F total. When we simplify this out is going to be five amps Times .8 m, I cross 0. or Tesla. Okay plus 0.4, - .4 m J Cross .34 Tesla. Yeah, okay. We can lug that in solve that out and we are left with - . Newton's I Direction -1. Newtons in the J Direction. So that gives us the force in terms of our components, but they're asking for the magnitude. So from there, all we have to do is recall Pythagorean theorem and we know that the magnitude is the square root of the sum of the squares. So that looks like . squared plus -1.3, 6 squared And that equals 1.5, 2 Newtons. So when we look at our potential answers can actually eliminate everything other than um other than D. But let's just Work through Part 2 to make sure that we're on the right track. Alright. So part two is asking for the angle. And so from our trigonometry, we can, if we know we have the opposite and the adjacent Sackatoga, we're working with our tangents. So our angle is going to be the inverse tangent of our opposite, which is RJ direction negative 1.36 over the adjacent - .68. And that equals degrees, Sorry 63.4°. We know that from the fact that we have the negative I and negative J terms for in quadrant three. And so from the positive X axis, it's actually this angle plus 100 and 80 degrees. And so that equals 243.4°. when measured from the positive x access, we go back up to our answers and yes, that is the correct answer for part two of choice D so D is the answer to this problem. That's all we have for this one. We'll see you in the next video.

Verified Solution

Key Concepts

Magnetic Force on a Current-Carrying Wire

Right-Hand Rule

Uniform Magnetic Field