Question

Regions 1 (z < 0) and 2 (z > 0) are separated at z = 0 by a current sheet of 3500 A/m in y-direction. Region 1, (z > 0, μr = 150) has a magnetic field intensity of 35?̂ + 55?̂ + 6?̂⁡ kA/m. If region 2 has a relative permeability of 650, calculate the magnetic field intensity in this region.

Answer 1

So, we can write magnetic field intensity at region 2 using boundary condition

B2n - B1n =0

here n represent normal conponent

so, normal component is along z direction so, H- field we write

μr2H2n - μr1 H1n=0

So, we have given μr2= 650 and μr1= 150

so we write

H2n = (150/650) 6? = 1.38 z

and tangential componenet is along x and y direction

so boundary condition

H1t - H2t = K

K is current along edge so because current flows along y direction

so,

H2y = 55 - 3.5 kA/m = 51.5 kA/m
and there is no current along x direction so.

H2x = H1x = 35

so we write H in region 2 as

H = 35?̂ + 51.5?̂ + 1.38?