A stiff wire 50.0 cm long is bent at a right angle in the midd...

Question

A stiff wire 50.0 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the line y = 2x in the xy plane. A current of 20.0 A flows in the wire—down the z axis and out the wire in the xy plane. The wire passes through a uniform magnetic field given by = (0.285î ) T. Determine the magnitude and direction of the total force on the wire.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with magnetic force. In this problem, a metal working shop constructs a device using 60.0 centimeters of rigid metal rod bent sharply at its B point to form a right angle. One section of the rod aligns with the X axis while the other extends into the YZ plane. Following the equation Z is equal to three Y. When testing the device, a 25.0 amp current passes through the rod flowing along the X axis and into the YZ plane. The entire assembly is placed in a uniform magnetic field described by B is equal to 0.350 Tesla J hat to validate the, the design determine the magnitude and direction of the total magnetic force on the rod. We're given four possible choices as our answers. For choice A, we have magnitude of 3.6 newtons and the direction is 100 and 33 degrees. For choice B, we have the magnitude of 6.2 newtons and the direction of 145 degrees. For choice C, we have the magnitude of 6.2 newtons and the direction of 100 67 degrees. And for choice D, we have the magnitude of 3.6 newtons and the direction of 100 70 degrees. So we need to calculate the force acting on this rod due to the magnetic field to recall your formula or the force acting on a current inside a magnetic field. So that is F is equal to IL cross B where F here is the force I is the current L is gonna be the length of the rod and B is our magnetic field. So our vector L in this case is going to be the length of the rod, but with the direction indicated by the direction the current is flowing through the rod. And so we'll need to write that out in component form. We have L is equal to the first part of the rod has the current flowing through half the rod that's aligned with the X axis. So our total length of the rod is 60 centimeters. So I'll convert that into meters by dividing by 100. So it's gonna be 0.6 m and I'll divide that by two to get it in half. And that um portion of the rod is going in the I hat direction since that is aligned with the rod. So I'll multiply that fraction by I hat. Then we're gonna have to add to that the other half of the rod which is gonna be 0.6 m divided by two. And that's gonna be multiplying the quantity. And here I have to um write the vector using this equation of Z is equal to three Y. So I'll have just my um J hat, which is gonna be my Y direction. And that means that our Z, which in this case is gonna be a K hat direction is gonna be three K hats. We'll have plus three K hat. So I've basically plugged in one for Y and into our equation. And I've gotten, I've gotten out three or RZ. Now, I want this to be a unit vector would divide that um J hat plus three K hat by its magnitude, which is going to be given by the square root of three squared plus one squared. And so I can just rearrange this um expression. So I'll have L is equal to, we'll have 0.6 divided by two in both terms. So I'll factor it out and go ahead and do that division. So I'll have 0.3 m and that's gonna be multiplying the quantity of I hat plus one divided by the square root of 10 J hat plus three, divided by the square root of 10 K hat. So now that we have an expression for our L vector, we can plug it in and actually calculate our force. So we're gonna have F it's equal to, for my current, I was given that in the problem, that is the 25 amps. And that's gonna be multiplying RL, which is the 0.3 m multiplying the quantity of I hat plus one divided by the square root of 10 J hat plus three, divided by the square root of 10-K hat. And that's gonna be crossed with our Magnan field which we were given in the problem, which is 0.35 Tesla J hat. So we'll try to simplify this a little bit. So we'll have F is equal to, I'm gonna multiply the 25 amps, the 0.3 m and the 0.35 Tesla together. And that's gonna give me 2.625 newtons and that's gonna be multiplying the quantity. Now, I can do the cross product, I'll have I hat cross J hat, which gives me K hat. Then the next term would be the J hat cross J hat, which is zero. And then I have the next term which is going to be the three divided by the square root of 10-K hat crossed with J hat. That's going to give me minus three divided by the square root of 10 I hat to recall that K hat cross J hat gives me minus I hat. So now I have my force vector here written in component form. I can go ahead and distribute the um 2.625 newtons. I'll have F is equal to 2.625 newtons, K hat minus 7.875 newtons divided by the square root of 10 I had. So the question wants us to find the magnitude and the direction of this force vector. So now I just need to calculate the magnitude. So for the magnitude of F that's gonna be equal to the square root of the quantity of 2.625 newtons squared plus the quantity of minus 7.875 newtons divide by the square root of 10 that Quintus squared. And so why I plug those values into my calculator, I get F is equal to 3.6 newtons. And here just kept two significant figures. That's one half of my answer. The other half of my answer is going to be the angle that the vector makes with the X axis. And so we'll label that as the, and that's gonna be equal to the arc tangent of. And here we'll have our K component in the numerator. So we'll have the 2.625 newtons divided by the minus 7.875 newtons divided by the square root of 10 and the denominator. But my vector, the F vector is located in the second quadrant of the XZ plane, but the arc tangent is only defined in the 1st and 4th quadrant. So we need to add 100 80 degrees to this angle in order to move it from the fourth quadrant to the second quadrant, which is where the vector is actually located. And so when I plug those values into my calculator, I'll get theta is equal to 100 and 33 degrees. And so that's the second half of my answer. And so now if I look at my answer choices, this actually corresponds to answer choice. A so just a quick recap of what we've done, we started off with our equation for the magnetic force acting on a current inside of a magnetic field. And in this case, we had to write our L vector in that formula using the fact that our rod was being bent at a 90 degrees. And the equation that Z is equal to three Y, which was given uh given us to us in the problem that told us where the second half of the rod actually lied. Once we wrote an expression for L, we just plugged that in and did the cross product. And eventually, once we have the uh force vector in component form, we just had to find the magnitude and direction using our normal uh methods for finding magnitude and direction of any vector. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Magnetic Force on a Current-Carrying Wire

Vector Components

Right-Hand Rule

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