Question

A long thin conducting sheet of width a carries a current I uniformly distributed across its width. The sheet lies in the XOY plane in the region a/2 ≤ y ≤ a/2, with the current flowing in the direction opposite to the x-axis.

          A long thin conducting sheet of width a carries a current I uniformly distributed across its width. The sheet lies in the XOY plane in the region a/2 ≤ y ≤ a/2, with the current flowing in the direction opposite to the x-axis.
        

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A long thin conducting sheet of width a carries a current I uniformly distributed across its width. The sheet lies in the XOY plane in the region a/2 ≤ y ≤ a/2, with the current flowing in the direction opposite to the x-axis.
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Transcript

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00:01 Hi there, so for this problem, we are told that a very large sheet of a conductor is located in the xy plane, as is shown in the figure, and it has an uniform current flowing in the y direction, and the density is given, and that is equal to 1 .5 ampers per centimeter.
00:32 So we need to use umpers law to calculate the direction in the magnitude of the magnetic field just above the center of the sheet.
00:45 So to find the magnetic field above the center of the surface of a current current sheet, we use umpers law.
00:57 So the path taken should be far from the edges and should be rectangular as shown in.
01:06 In this diagram that we are going to choose.
01:11 So we have some in here, some magnetic field in here.
01:17 So are you going to choose the ones in the middle? we're going to choose this one in here.
01:30 And we will have some path in here.
01:46 So we have the magnetic field b1, the magnetic field b2, so the direction of the magnetic field is found using the right -hand rule to be in the adds positive direction about the surface of the conductor.
02:10 So on per's law states that the close integral of the product between the magnetic field and the bed door of length is equal to the magnetic field times two times pi times the radius is equal to new sub zero times the current enclose so as you can see we're gonna label this we're gonna set that this is one two three and four so we can know that exceptions 1 and 3 are perpendicular to the films.
03:01 So from that we have that the magnetic field times the ds is going to be equal to 0 because those are perpendicular.
03:14 As you can see, the magnetic field is in the paper...
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