Find the magnetic field of an infinite uniform surface

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Homework Statement


234234-png.90190.png

This example is from 3rd edition of Griffiths' textbook. Ex. 5.8 on page 226
I understand that by reversing the direction of the current, sign of B is switched. but i can't get it that highlighted part. and why B doesn't have z-component???

Homework Equations




The Attempt at a Solution

 

Answers and Replies

  • #2
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7

Homework Statement


234234-png.90190.png

This example is from 3rd edition of Griffiths' textbook. Ex. 5.8 on page 226
I understand that by reversing the direction of the current, sign of B is switched. but i can't get it that highlighted part. and why B doesn't have z-component???

Homework Equations




The Attempt at a Solution

Think about what would happen if we rotated the coordinate system. Originally the current was in the +x direction, but we rotate our original coordinate system -90 degrees around the z-axis so that now the flow of K is in the +y direction. The z-component of B wouldn't change, correct?

Now if we changed the direction of the flow of current in the plane, it is analogous to the rotation above. The z-component of B can't change by simply changing where we define x and y. Therefore there is no z-component of B.
 
  • #3
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Think about what would happen if we rotated the coordinate system. Originally the current was in the +x direction, but we rotate our original coordinate system -90 degrees around the z-axis so that now the flow of K is in the +y direction. The z-component of B wouldn't change, correct?

Now if we changed the direction of the flow of current in the plane, it is analogous to the rotation above. The z-component of B can't change by simply changing where we define x and y. Therefore there is no z-component of B.


2342323.PNG


But the current flow in the infinite plane. if i draw a lot of loops as seen in the picture,
i think B is canceled out and entire B can not be existed , what is wrong with my thought??
 
  • #4
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The magnetic field beneath the plane of current travels in the opposite direction of that above the plane (use the right hand rule... above the plane, B is in the -y direction, and below the plane, B is in the +y direction). Thus when using Biot-Savart Law, the integral around the loop is the sum of the field above the plane times the length l plus the field below the plane times l (the vertical edges have no contribution). Notice that both above the plane and below the plane, the magnetic field B is in the same direction as the path ds. Therefore the fields will not cancel; they will add.
 
  • #5
50
0
The magnetic field beneath the plane of current travels in the opposite direction of that above the plane (use the right hand rule... above the plane, B is in the -y direction, and below the plane, B is in the +y direction). Thus when using Biot-Savart Law, the integral around the loop is the sum of the field above the plane times the length l plus the field below the plane times l (the vertical edges have no contribution). Notice that both above the plane and below the plane, the magnetic field B is in the same direction as the path ds. Therefore the fields will not cancel; they will add.


thank you. i got it !
 

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