Question

Two regions of magnetic materials are separated by the z=0 plane. In the region with z>0, the magnetic field intensity is given by H = 40 + 50x + 122kA/m. The relative permeability in this region is 20. a. (5 pts) Determine the magnetic field intensity in the region with z=0, which has a relative permeability of 100. Take into account a surface current density of J = 10kA/m at z=0. b. (4 pts) Calculate the bound surface charge density and determine whether the materials are homogeneous. Attach File: BranzeMyCompute

          Two regions of magnetic materials are separated by the z=0 plane. In the region with z>0, the magnetic field intensity is given by H = 40 + 50x + 122kA/m. The relative permeability in this region is 20.

a. (5 pts) Determine the magnetic field intensity in the region with z=0, which has a relative permeability of 100. Take into account a surface current density of J = 10kA/m at z=0.

b. (4 pts) Calculate the bound surface charge density and determine whether the materials are homogeneous.

Attach File: BranzeMyCompute
        
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two regions of magnetic materials are separated by the z0 planein the region with 0the magnetic field intensity is given by 4050x122ka permthe relative permeabity inthis region is20 a5 ptsde 78843

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Two regions of magnetic materials are separated by the z=0 plane. In the region with z>0, the magnetic field intensity is given by H = 40 + 50x + 122kA/m. The relative permeability in this region is 20. a. (5 pts) Determine the magnetic field intensity in the region with z=0, which has a relative permeability of 100. Take into account a surface current density of J = 10kA/m at z=0. b. (4 pts) Calculate the bound surface charge density and determine whether the materials are homogeneous. Attach File: BranzeMyCompute
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Transcript

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00:01 Hello students here we have the magnetic field b vector equal to 3x a x vector plus minus 3 x square y plus 9 x square a y vector plus 2 y minus 3 z plus 3 x square z a z vector.
00:13 So here we have to calculate the magnetic flux.
00:15 So magnetic flux can be determined over a surface as phi equal to double integral over a surface into b d s vector.
00:23 So here we can write down the s vector as s x equal to x comma x square plus y square because we have the parabola surface here y equal to x square plus z square.
00:36 So substituting it at calculating the derivatives dou s by dou x and dou y by dou z that is 1 comma 2 x comma 0 and 0 comma 2 z comma 1.
00:47 So from this we can get the normal vector as dou s divided by dou x cross dou s divided by dou z.
00:53 So we will get the cross product as 2 x comma minus 1 minus 2 z.
00:59 Then we can get the normal vectors amplitude magnitude as square root of 4 x square plus 1 plus 4 z square.
01:05 So the unit normal is the vector divided by its magnitude and then we can get the d s vector as cross product between dou s by dou x and dou s divided by dou y into dx dz.
01:19 So we will get d s vector as 2 x a y minus i a y plus 2 z a z magnitude into dx dz...
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