## $$-\left(2 I B_{0} r \sin \theta_{0}\right ) j$$

### Discussion

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##### Christina K.

Rutgers, The State University of New Jersey

##### Marshall S.

University of Washington

### Video Transcript

case we're doing Chapter 27 problem. So we have a current carrying circular loop of wire with radius are current, and currents go. Traveling counterclockwise is shown here in the next life. Um, it says it's partially immersed in the magnetic fuel. Constant magnitude Be not directed out of the page. Okay, So determined the net force on the loop to to the field in terms of st Amant. Okay, cool. So first, we're gonna find the net force on the loop by integrating the infant testable force on each incontestable portion of the with a wire. So the infinitesimal force equation found Equation 27 4 to 1 on the back. So we have i d l is our custom all portion of the loop here, huh? And, um, this pair of a tries for a circle circular loop can just be easily written as bye terms negative coastline, theater and the eyes direction plus groups cover waas, Sinus ada in the J direction times the radius, do you think? Okay, so now we're going to use this our force equation. We know that the forces getting by the integral i d o cross b. Okay, so now If we plug in what we have here, we easily conceive that this becomes integral from negative co sign. I, uh, current out in front. What Sign? They jihad. Okay, this is sometimes our defeat, huh? And we cross this with R B, but we can actually write be even easier than this. So instead of writing the we know that it's be not, which is the magnitude. And, you know, it's a big, positive K direction out of the page so we can write that out like this. Okay, so the next part is we need to figure out our bounds. So if the entire loop was immersed in our magnetic field to be 11 of the boot we want wanna integrate data from zero all the way to two pi, but we don't want to do that because the whole field is not in the loop or not. The whole loop is not in the field. So this whole section up here has no magnetic field. So really, we need to start from fate or not and integrate all the way out here. This angle, which is to pi minus stated on so are integral, becomes starting from state and not integrating to two pi minus state or not cool. So if that's our integral, let's now evaluate the cross product before we evaluate the integral and we see the forced outcomes I be not are in a girl from they don't not to pi minus state or not. And now, evaluating this cross product of these angles across with Jay because hosts they, uh J huh? Because sorry, the and this is all immigrated from deep data. Awesome. So now we can just easily do this in a girl because we know the integral of coast. They don't know the integral sign data. So let's evaluate this and you get I be not r and that we evaluate. So hiring the, uh j hi minus coast I had and this all is evaluated from our balance. They did not thio two pied by mistake. So now we're just gonna put this stuff in and you see this consent to being I do not r you know, sign of loose, not it's good luck. Sign of two pi minus state or not. J hat. Finest sign. They did not J huh? This co sign to pi minus state or not? I, uh, plus co sign. They do not. I cool. Then we should see that these things are gonna cancel out, because the, um, coastline of two pi minus state or not is exactly the opposite of for the same with Poseidon State or not. So these air can slough. And this is gonna cancel out as well because, you know, all right, please, I didn't do that to pi minus did not equals negative sign. They do not and co sign of two pi minus state. Not he calls saying was kosher. So I mean, these are gonna cancel out. And these are gonna add to each other such that our final answer for our force here is given by negative to I you know, or sign. They don't know J. And that's the answer.

University of Oregon
##### Christina K.

Rutgers, The State University of New Jersey

##### Marshall S.

University of Washington