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Boundary μr = 10, H given in region 1

Short answer

Tangential H continuous: H2x=5, H2z=7 A/m. Normal B continuous: μ0*6 = 10μ0*H2y, so H2y=0.6 A/m. H2 = 5a_x + 0.6a_y + 7a_z A/m. B2 = μ0*10*H2 = 62.8a_x + 7.54a_y + 88.0a_z μT.

The region y < 0 (region 1) is air and y > 0 (region 2) has μ_r = 10. If there is a uniform magnetic field H = 5a_x + 6a_y + 7a_z A/m in region 1, find B and H in region 2.

Region 1 (y < 0) is air (μ_r1 = 1) and region 2 (y > 0) has μ_r2 = 10. The interface is the plane y = 0, so the normal is a_y and the tangential components are x and z. There is no surface current stated (K = 0).

H₁ = 5 a_x + 6 a_y + 7 a_z A/m.

Step 1: B in region 1 (for reference).

B₁ = μ₀ H₁ = 4π×10⁻⁷ (5, 6, 7) = 6.283 a_x + 7.540 a_y + 8.796 a_z  μT

Step 2: boundary conditions.

  • Tangential H is continuous (K = 0): H₂x = 5, H₂z = 7 A/m.
  • Normal B is continuous: μ₁H₁y = μ₂H₂y
H₂y = (μ₀ × 6) / (10 μ₀) = 0.6 A/m

Step 3: H in region 2.

H₂ = 5 a_x + 0.6 a_y + 7 a_z   A/m

Step 4: B in region 2.

B₂ = μ₀ μ_r2 H₂ = 4π×10⁻⁷ × 10 × (5, 0.6, 7)
   = 62.83 a_x + 7.54 a_y + 87.96 a_z   μT

Answer: H₂ = 5 a_x + 0.6 a_y + 7 a_z A/m; B₂ = 62.8 a_x + 7.54 a_y + 88.0 a_z μT (note B_y is unchanged at 7.54 μT).

✓ Answer checked by Claude Opus 5.5 (high effort)

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