Let there be one sheet of infinite length and breadth is shown in the figure given below. Electric current is flowing along the length of sheet (perpendicular to the plane of paper). K is the electric current per unit width in this given sheet. Evaluate the magnetic field intensity near this sheet of current.

Explanation

Refer the above given figure to understand that the shown length and breadth of sheet are infinite and its length is perpendicular to the plane of paper. A negligible width of sheet is also shown in the figure and the sheet carries outward flow of current .
We can visualise this sheet as several infinitely long wires kept parallel to each other in contact. Let us select the width dl of sheet then it is equivalent to an infinitely long wire that is kept perpendicular to the plane of paper carrying current Kdl. The direction of current in the figure is shown to be outward.
Refer the below given figure. Let there be a point P somewhere close to this sheet of current where we need to determine the magnetic field due to the sheet. Let us drop a perpendicular on the sheet from this point P and let M be the foot of this perpendicular.

If we select two segments symmetrically on the two sides of point M, then we can easily understand that the resultant magnetic field due to these two segments will be parallel to the sheet as shown in the figure. Hence roughly, magnetic lines are, as shown, around the sheet of current .
Now select rectangular Amperian loop PQRS as shown in the following figure. Sense of the loop is as shown in the figure.

RS and PQ are equidistant from the sheet of current so the magnitude of magnetic field intensity on these two sides will be same and let it be B. The remaining two sides are perpendicular to the field of sheet. We can see that the total circulation of magnetic field for this loop will be 2BL.
According to Ampere.s law, the total circulation of magnetic field must be equal to `mu_(0)` times the enclosed current. Thus, we can write the following:
`ointvecB*vec(dl) = mu_(0)i`
` rArr" " 2BL = mu_(0)(KL)`
` rArr" " B = (mu_(0)K)/2 `