# Find the magnetic field of an infinite uniform surface

## Homework Statement

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???

## The Attempt at a Solution

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## Homework Statement

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???

## 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.

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.

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??

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.

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 !