A wire of \(3 \mathrm{~mm}\) radius is made up of an inner material \((0<\rho<2 \mathrm{~mm})\) for which \(\sigma=10^{7} \mathrm{~S} / \mathrm{m}\), and an outer material ( \(2 \mathrm{~mm}<\rho<3 \mathrm{~mm}\) ) for which \(\sigma=4 \times 10^{7} \mathrm{~S} / \mathrm{m}\). If the wire carries a total current of \(100 \mathrm{~mA}\) dc, determine \(\mathbf{H}\) everywhere as a function of \(\rho\).

First, we need to determine the distribution of the total current in both the inner and outer materials of the wire. We know the total current is 100 mA and the conductivities of the materials are σ₁ = 10^7 S/m and σ₂ = 4×10^7 S/m. The current density J is the current per unit area and can be related to the conductivity and the electric field as follows: J = σE Since the electric field E is constant within each cylindrical material, their current densities J₁ and J₂ inside each material are also constant. Let the current in the inner material be I₁ and the current in the outer material be I₂. We can express the total current I (100 mA) in terms of I₁ and I₂: I = I₁ + I₂ Then, we can write the current densities J₁ and J₂ in terms of I₁ and I₂: J₁ = I₁ / A₁ J₂ = I₂ / A₂ Where A₁ and A₂ are the cross-sectional areas of the inner and outer materials. A₁ = π(2 mm)² A₂ = π[(3 mm)² - (2 mm)²] Now, we can use the proportions of conductivities to find the relation between I₁ and I₂: J₁ / J₂ = σ₁ / σ₂ (I₁ / A₁) / (I₂ / A₂) = σ₁ / σ₂

Now we have two equations with two unknowns (I₁ and I₂), which we can express as follows: I₁ + I₂ = 100 mA (I₁ / A₁) / (I₂ / A₂) = σ₁ / σ₂ Solve these linear equations simultaneously to find I₁ and I₂.

For an Azimuthal magnetic field \(\textbf{H}\) in a cylindrical coordinate system, Ampere's circuital law is given as: ∮H dL =NI For the inner region (0 < ρ < 2 mm), we have I = I₁, and for the outer region (2 mm < ρ < 3 mm), I = I₃.

Now we can calculate the magnetic field intensity H as a function of ρ for both the inner and outer regions individually. For the inner region (0 < ρ < 2 mm): H(ρ) = I₁ / (2πρ) For the outer region (2 mm < ρ < 3 mm): H(ρ) = I₂ / (2πρ)

Combine the expressions for the magnetic field intensity H(ρ) in both regions into a single expression: H(ρ) = { I₁ / (2πρ) if 0 < ρ < 2 mm , I₂ / (2πρ) if 2 mm < ρ < 3 mm } This is the magnetic field intensity H(ρ) everywhere as a function of ρ for the given wire.