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(III) A curved wire, connecting two points a and b, lies in a

plane perpendicular to a uniform magnetic field $\vec{\mathbf{B}}$ and

carries a current $I .$ Show that the resultant magnetic force on the wire, no matter what its

shape, is the same as that on a

straight wire connecting the two

points carrying the same current I.

See Fig. $40 .$

$I B_{0}(-\Delta x \hat{\mathbf{x}}+\Delta y \hat{\mathbf{i}})$

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Okay, so we're gonna Chapter 27 problem 11. So it says a curved wire connecting between two points and lies and a plane perpendicular to remain attics. He'll be. So we have a shot into the page here, and the wire carries a current I So it says show that the resultant magnetic force on the wire, no matter what its shape is, is one travels from A to B is the same as that on a straight wire connecting the two points. A to be hearing the same weight. Same couple. Okay, so first, how we can start this problem is using equation 27 4 We want to integrate to find the in for testing off force along each path element along an arbitrary passed between a beat. So that sounds confusing. So if you look back at Equation 27 before we can write the force here as the interval from some path A to B i d. L cross B. And this is where d l um talks about our path here. So we can now, right? D l Then that's just the back to her, pointing from eight b and this does completely independent of the path as long as we have the same current on these wives. So if we write out what r d l is, then we haven't vector d'oh! Where this is equal to i d x plus j Why, Okay, So if we plug that into her in Nicole, we now see that this becomes I and a girl from a to B Oh, I d x plus J d y addresses crossed books, cross being our terms. Okay, Okay. So now if we just do evaluate this integral or value it across products and Indian girl see that this comes out to be I be not a to b of negative j t X plus I buy. So now if we just integrate this this now becomes I You know, uh, and now we have negative X j Plus, that's why uh, Mrs Oliver from a to B So now we pull again. These points a b we see that this is I be not Now we have negative b x j waas Who's must be why All right. Waas a X J linus, There you are. And you should see that this just becomes the delta. Are the change and potential are changing positions. Right, So B x plus B R B x minus B Y here is just sorry. Be hex minus a X here gives us a negative for Delta tax. And we have a delta y from the difference here. So that's how about I be now and then we have negative Delta X and E J direction. Plus don't ally in the IRS. So this was the key here. We get to this point and we see that our answer force Onley depends on the change from the two points, right? This depends on the difference. The displacement between points A and B, But it doesn't depend on the path because nothing we plugged in here actually depends on the path we took. That's the key. So the force on this wire, as long as it's traveling from point A to B and has the same white current, I will always be this value so that the grid really keep Keith Point taken Take for this problem