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Boundary condition with μr = 10, find B and H

Short answer

H₂ = 5 a_x + 6 a_y + 7 a_z A/m (given). B₂ = μ₀ μ_r2 H₂ = (4π×10⁻⁷)(10)H₂ = 62.8 a_x + 75.4 a_y + 88.0 a_z μT (≈ 6.28×10⁻⁵ a_x + 7.54×10⁻⁵ a_y + 8.80×10⁻⁵ a_z T).

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 2, find B and H in region 2.

Region 1 (y < 0) is air (μ_r1 = 1). Region 2 (y > 0) has μ_r2 = 10. The boundary is the plane y = 0 with normal a_y. The field given is in region 2: H₂ = 5 a_x + 6 a_y + 7 a_z A/m.

H in region 2

The field H₂ is given directly: H₂ = 5 a_x + 6 a_y + 7 a_z A/m, with |H₂| = √(25+36+49) = 10.49 A/m.

B in region 2

B₂ = μ₀ μ_r2 H₂ = (4π×10⁻⁷)(10)(5 a_x + 6 a_y + 7 a_z)
   = 1.2566×10⁻⁵ (5 a_x + 6 a_y + 7 a_z)
   = 62.8 a_x + 75.4 a_y + 88.0 a_z  μT   (μWb/m²)

In tesla: B₂ = 6.28×10⁻⁵ a_x + 7.54×10⁻⁵ a_y + 8.80×10⁻⁵ a_z T.

Corresponding fields in region 1 (using boundary conditions)

No surface current is stated, so tangential H is continuous and normal B is continuous:

H_t1 = H_t2 = 5 a_x + 7 a_z
B_N1 = B_N2 :  μ₀ H_y1 = 10 μ₀ (6)  →  H_y1 = 60
H₁ = 5 a_x + 60 a_y + 7 a_z  A/m
B₁ = μ₀H₁ = 6.28 a_x + 75.4 a_y + 8.80 a_z  μT

Answer: H₂ = 5 a_x + 6 a_y + 7 a_z A/m; B₂ = 62.8 a_x + 75.4 a_y + 88.0 a_z μT. (Region 1: H₁ = 5 a_x + 60 a_y + 7 a_z A/m.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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