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All right, we're looking at question 28 .34.
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And we have this semicircular current of wire.
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And we're looking at point p, and we want to find the magnitude and direction of the magnetic field at point p.
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So we're going to need to use the b .s of our law, which i've written down up here.
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And the key components that we have, here's the constant out in front.
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I is the current.
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Dl is going to be the little section of wire, the segment that we're looking at along the length of the wire.
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And sine of phi, the angle, is going to be the angle between r, which is going to be the vector pointing to the point that we're looking at.
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That's going to be r here and dl.
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So first let's think about these straight sections of wire here.
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We have, so this term right over here where we have i times dl times r times the sign of this angle, this is really a cross product.
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And the cross product is between dl and r.
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So we want to look at these two red vectors here.
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So when we're looking at these straight sections, we're going to have dl is just going to be, along just a horizontal line just like this.
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So this is going to be our dl and our r is going to be just pointing directly horizontally out in the other direction.
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So our dl and an r are going to be parallel and the cross product between two parallel lines is going to be zero.
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And the same situation is going to happen over here.
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The dl and the r for the straight section of wire.
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They're both parallel so the cross product is zero.
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So the these straight segments are not going to contribute anything to the magnetic field at point p.
02:01
So we're just going to totally ignore them.
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Now this semicircular section here, when we look at dl at any point, it's going to be the tangent to the semicircle.
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And then r is going to be along the radius of the circle.
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So dl and r are going to be perpendicular at every point in the semicircle.
02:24
So sine of theta or sine of phi here when we're saying that we're looking at this angle between dl and r, sine of theta is really just going to be one because sine of 90 degrees is one.
02:41
So we can actually just get rid of this angle term here and not worry about it because it's just going to be equal to one for the entire integral that we have to do here.
02:53
So let's just plug these in in terms of the things, the terms that we know.
02:59
So we know that our little r here is going to be equal to the radius of the semicircle, which is just capital r.
03:07
And we have the current.
03:10
I guess the current is just listed as i.
03:13
So let's just rewrite this just using our capital r just to be totally explicit.
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It...