Ampere's law and boundary H components

Ampere's circuital law: the line integral of H around any closed path equals the current enclosed by the path, ∮H·dl = I_enc (point form ∇×H = J).

Boundary conditions used

The boundary is the plane x = 0, so the normal is a_x and tangential components are the a_y and a_z components. Region A (x < 0): μ_A = 5 μH/m. Region B (x > 0): μ_B = 20 μH/m. K = 150 a_y − 200 a_z A/m. H_A = 300 a_x − 400 a_y + 500 a_z A/m.

• Normal B is continuous: μ_A H_NA = μ_B H_NB.
• Tangential H jumps by the surface current: a_N12 × (H_B − H_A) = K, with a_N12 = a_x pointing from A to B.

(a) |H_LA| (tangential component in A)

H_LA = −400 a_y + 500 a_z
|H_LA| = √(400² + 500²) = √410000 = 640.3 A/m

(b) |H_NA| (normal component in A)

H_NA = 300 a_x ,  |H_NA| = 300 A/m

(d) |H_NB| (normal component in B)

μ_B H_NB = μ_A H_NA
H_NB = (5/20)(300) = 75 a_x ,  |H_NB| = 75 A/m

(c) |H_LB| (tangential component in B)

Let H_LB − H_LA = ΔH_y a_y + ΔH_z a_z. Then

a_x × (ΔH_y a_y + ΔH_z a_z) = ΔH_y a_z − ΔH_z a_y = 150 a_y − 200 a_z
→ ΔH_z = −150 ,  ΔH_y = −200

H_LB = (−400 − 200) a_y + (500 − 150) a_z = −600 a_y + 350 a_z
|H_LB| = √(600² + 350²) = √482500 = 694.6 A/m

Result

Hence H_B = 75 a_x − 600 a_y + 350 a_z A/m.