Boundary at z=0, find B2

The boundary is the plane z = 0, so the normal is a_z and the tangential components are the x and y components. Region 1 (z > 0, μ₁ = 4 μH/m) is above, region 2 (z < 0, μ₂ = 7 μH/m) is below. A surface current K = 80 a_x A/m flows on the boundary. The normal pointing from region 2 to region 1 is a_N12 = a_z.

B₁ = 2 a_x − 3 a_y + 1 a_z mT.

Step 1: normal component (B continuous).

B₂z = B₁z = 1 mT

Step 2: tangential H in region 1.

H₁t = B₁t / μ₁
    = (2 a_x − 3 a_y)×10⁻³ / (4×10⁻⁶)
    = 500 a_x − 750 a_y   A/m

Step 3: apply the boundary condition a_N12 × (H₁ − H₂) = K. Let H₁t − H₂t = h_x a_x + h_y a_y.

a_z × (h_x a_x + h_y a_y) = h_x a_y − h_y a_x = 80 a_x
⇒ h_x = 0,  h_y = −80

So H₂t = H₁t − (0 a_x − 80 a_y):

H₂t = 500 a_x + (−750 + 80) a_y = 500 a_x − 670 a_y   A/m

Step 4: tangential B in region 2.

B₂t = μ₂ H₂t = 7×10⁻⁶ (500 a_x − 670 a_y)
    = 3.5×10⁻³ a_x − 4.69×10⁻³ a_y   T

Step 5: combine.

Answer: B₂ = 3.5 a_x − 4.69 a_y + 1 a_z mT