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christinaaa.g
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Hi everyone,
1.
Part (a)
Find the magnetic field above and below an infinite current sheet (infinite straight wires with current I along the positive z-axis). The sheet is on the xz plane. There are N wires per meter counting along the axis.
Part (b)
There are two parallel sheets with the same givens as part (a) except that the second sheet is separated by a distance d above the first. The sheet above carries current I along the positive z-axis. Find the magnetic field above, between, and below the sheets.
2.
Ampere's Law:
Integral of (gradient cross B) times dA=Integral of B times dl
with B=magentic field
A=area
l=length
3. I'm having more trouble with part (b) than I did with part (a).
This is how I went about solving part (a). I'm not sure if it's completely right or not though.
I drew a rectangular loop parallel to the xy plane extended an equal distance above and below the surface. I then used ampere's law...
Integral (B times dl) = 2Bl
It is 2Bl because one is from the top segment and one from the bottom.
2Bl=(mu sub 0)(I enclosed)(l)
where I is the current and l is the length and mu is the permeability of free space
Setting 2Bl = 0(I enclosed l)
B=+ (mu sub 0/2)I in the x direction (when y<0)
B= - (mu sub 0/2)I in the x direction (when y>0)
If anybody has any ideas to solve part (b) or any corrections to my part (a) solution, I would really appreciate it! Thanks!
1.
Part (a)
Find the magnetic field above and below an infinite current sheet (infinite straight wires with current I along the positive z-axis). The sheet is on the xz plane. There are N wires per meter counting along the axis.
Part (b)
There are two parallel sheets with the same givens as part (a) except that the second sheet is separated by a distance d above the first. The sheet above carries current I along the positive z-axis. Find the magnetic field above, between, and below the sheets.
2.
Ampere's Law:
Integral of (gradient cross B) times dA=Integral of B times dl
with B=magentic field
A=area
l=length
3. I'm having more trouble with part (b) than I did with part (a).
This is how I went about solving part (a). I'm not sure if it's completely right or not though.
I drew a rectangular loop parallel to the xy plane extended an equal distance above and below the surface. I then used ampere's law...
Integral (B times dl) = 2Bl
It is 2Bl because one is from the top segment and one from the bottom.
2Bl=(mu sub 0)(I enclosed)(l)
where I is the current and l is the length and mu is the permeability of free space
Setting 2Bl = 0(I enclosed l)
B=+ (mu sub 0/2)I in the x direction (when y<0)
B= - (mu sub 0/2)I in the x direction (when y>0)
If anybody has any ideas to solve part (b) or any corrections to my part (a) solution, I would really appreciate it! Thanks!