Electromagnetics Test 2

The electric field phasor w.r.t distance ‘y’ is given, time dependence of electric field must satisfy the Maxwell equation.

The given equation is similar to phasor of uniform plane wave in free space

\( \Rightarrow \frac{\omega }{\beta } = c\)

ω = 4 × 3 × 10⁸ rad/s

A static electric field in a charge-free region is defined as

∇ ⋅ E = 0

And ∇ × E = 0

(a → 4)

A static electric field in a charge region satisfies:

\(\nabla \cdot E = \frac{{{\rho _v}}}{\varepsilon } \ne 0\)

And ∇ × E = 0

(b → 2)

A steady magnetic field in a current-carrying conductor satisfies:

∇ ⋅ B = 0

∇ × B = μ0J ≠ 0

(c → 1)

A time-varying electric field in a charged medium with time-varying magnetic field has:

\(\nabla \times E = - \frac{{\partial B}}{{\partial t}} \ne 0\)

\(\nabla. \cdot E = \frac{{{\rho _v}}}{\varepsilon } \ne 0\)

(d → 3)

Conduction current density J_C = σE

= 10 × 200 sin(ωt)

Displacement current density is given by:

\({J_d} = \frac{{\partial D}}{{\partial t}}\)

= ϵ₁ϵ₀ × 200 × ω cos(ωt)

∴ The frequency at which both currents are equal will be:

10 × 200 = ϵ_rϵ₀ × 200 × ω

\(f = \frac{\omega }{{2\pi }} = \frac{{10}}{{2 \times \varepsilon \times 2 \times 8.854 \times {{10}^{ - 12}}}}\)

The electric flux density is given by:

\(\vec D = \epsilon\vec E \)

Since \(E= \frac{{ V}}{d}\), the above expression becomes:

\(D= \frac{{\epsilon V}}{d}\)

Also, the displacement current density is defined as:

\({{\vec J}_d} = \frac{{\partial \vec D}}{{\partial t}} = \frac{\epsilon}{d}\frac{{\partial V}}{{\partial t}} \)

\( {I_d} = {J_d}.A = \frac{{\epsilon A}}{d}\frac{{dV}}{{\partial t}}\)

Calculation:

Putting on the respective values, we get:

\({I_d} = \frac{{{{10}^{ - 9}}}}{{36\pi }} \times \frac{{6 \times {{10}^{ - 4}}}}{{6 \times {{10}^{ - 3}}}} \times {10^3} \times 36\cos {10^3}t \)

\(= \frac{{100}}{\pi }\cos {10^3}t\;nA \)

Concept:

The boundary condition for electrostatic fields are defined as:

E_1t = E_2t (Tangential components are equal across the boundary surface)

Also, the normal component satisfies the following relation:

D_1n – D_2n = ρs

For a charge-free boundary, ρs = 0, the above expression becomes:

D_1n – D_2n = 0

D_1n = D_2n

Analysis:

Given, the electric field intensity in medium 1.

E₁ = 5a_x – 2a_y + 3a_z

Since, the medium interface lies in plane z = 0

So, we get the field components as

E_1t = 5a_x – 2a_y

And E_1n = 3a_z

Now, From the boundary condition for the electric field we have:

E_1t = E_2t

ε₁E_1n = ε₂E_2n

\({E_{2n}} = \frac{{{\varepsilon _1}}}{{{\varepsilon _2}}}{E_{1n}} = 6{a_z}\)

So, the field components in medium 2 are:

E_2t = E_1t = 5a_x – 2a_y

E_2n = 6a_z

Therefore, the net electric field intensity in medium 2 will be:

E₂ = E_2t + E_2n = 5a_x – 2a_y + 6a_z

So, the z-component of the field intensity in medium 2 is

E_2z = 6a_z