Question

Two Infinite thin sheets of current are parallel to each other. Both sheets conduct a linear current density of 8 A/m, but the top sheet current is out of the page, while the bottom sheet carries current into the page:

A. Calculate the magnetic field at point B, halfway between these sheets.

B. Calculate the magnetic field at point A, 1 meter above the plate.

C. The bottom plate current is changed so that a magnetic field at point C is measured to be 3uT to the right. If the top current density remains the same, what is the new current density in the bottom plate? Is it still inward?

1. Two infinite thin sheets of current are parallel to each other. Both sheets conduct a linear current density of 8 A/m, but the top sheet current is out of the page, while the bottom sheet carries current into the page: A 8 A/m (out) B 8 A/m (in) A. Calculate the magnetic field at point 'B', halfway between these sheets. (10 JT, right) B. Calculate the magnetic field at a point 'A', 1 meter above the top plate. C. The bottom plate current is changed so that a magnetic field at point 'C' is measured to be 3 uT to the right. If the top current density remains the same, what is the new current density in the bottom plate? Is it still inward? (3.2 A/m, in)

Answer 1

--- I cout) ☺ -.- for an infinite thin: sheet carring current density I, the magnitude of magnetic field above the Plate is, 1 Bir a Moth a direction is towards left & the magnetic field below the sheet is also C Bb = I a 2 towards night. is but direction found by night direction hand can be senew pule.

- - 8 Alm cout ) Beton - . n&am cin) 8 An . Now, magnetic field for two such sheet would be the Super position of magnetic fields due to each of the sheet. A) the magnetic field due to the upper Plate , ut position B "s, BF Mof which is towards า พ. since the bottom sheet is canning the current directed into the page,

hence, magnetic to the botton field Plate cut B, due is, - B₂ = M. I a the direction is 2 towards so, net night. magnetic field at point B ?d, B = BitB2 - Mof that = 2) B a Mol B = 40x16? x 8 T + B = 4ax.8 MT = B = 10.0530964915 MT B = 10 MT (appoon/ & the direction is towards night.

B) at point a, the magnetic field due to the upper Plate - B. = Mol e the direction is towards left, & the magnetic field due to the bottom plate is B2 = ". I & the direction is towards night. Hence they cancel each other. So, nesultant magnetic field at Point A is zeno. - given the magnetic field at point c .is, - Be=3ut e the direction is towards night.. . Now, the current through the bottom plute has been changed, but through the top plate it is Sare.

the Now, due to magnitude of ID, B = M. I single sheet magnetic field 2 > B = 40x10?xa T B = 5.02 65 482A5J MT so, the current through the botton Plate should be into the pase such that, the magnetic field due to it at point e, has direction towards left. density let the current through the botton Plate is Ib. so, we have, In me = 32 106 >> 4011528 my za a 331054 es ? Jy mom

=> Jy = 23(4- ) Alm → Io = (8- ) Ain 2) Ib 2. 3.225 3517072 Alm a) Io = 3.2 Alam & the direction is into the page.