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hd28cw
Dec7-04, 08:33 PM
Find the magnetic flux crossing the portion of the plane
theta = x/4 defined by 0.01 m < r <0.05 m and 0 m < z < 2 m. A current of 2.50 A is flowing along z-axis along a very long wire.

in drawing the picture i know that there is a long thin wire with a current of 2.5 amps flowing positively on the z-axis and the plane is lying rotated at an angle of pi/4 with the magnetic field flowing in a counter clockwise direction.

How do I go about finding the magnetic flux.
Please help.

Tide
Dec7-04, 10:40 PM
The magnetic flux through the surface is the integral

\int_S \vec B \cdot \hat n dA

where \hat n is a unit vector normal to the element of area dA.

From your description I don't have a clear picture of the geometry.

dextercioby
Dec7-04, 10:48 PM
The magnetic flux through the surface is the integral

\int_S \vec B \cdot \hat n dA

where \hat n is a unit vector normal to the element of area dA.

From your description I don't have a clear picture of the geometry.
Me neither.I saw this problem 2 hours ago,but i couldnt' figure out the geometry and how would flux definiton and Ampère's law fit in.Let's hope (for his sake) he comes up with a drawing or with the original text.

EDIT:Maybe i got it.He's obviously using cylindrical coordinates and that domain should be a rectangle perpendicular to the OXY plane with one of the sides lying in the OXY plane.So i guess,be should find that,since magnetic field lines are perpendicular to the rectangle,the angle between the magnetic field and \vec<n should be zero.At constant z,the field is varying inversly proportional with "r" as stated by Ampèere's law.So the surface integral would splin in 2,the integral along the "z" gives the magnitude of "z" and along "r" smth about natural logarithm.
I guess that does it.Yet where does the XY angle "theta" come in...??Or am i missing something??Like a different from pi over 2 angle between teh field and the rectangle??Anyway...Tide,i let u give your opinion.

Tide
Dec7-04, 11:25 PM
Dexter,

That makes sense. I think I'll wait for hd28cw to offer clarification!