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

Observation:

The normal component of electric flux density is continuous, but the normal component of the electric field will be discontinuous and inequal.

Analysis:

According to the boundary conditions:

E_1t = E_2t = E_3t    ---(1)

Also for charge free interface, the normal components will be related as:

ϵ₁ E_1n = ϵ₂ E_2n = ϵ₃ E_3n

Since ϵ₁ = ϵ₃

E_1n = E_3n ≠ E_2n    ---(2)

As the net electric field is E = Et + En, using equation (1) and (2), we can conclude that:

E1 = E3 ≠ E2    

∴ Options 2, 3, and 4 are correct.

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 = ρ_s

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

D_1n – D_2n = 0

D_1n = D_2n

Analysis:

Consider the two dielectric regions as shown below:

Since the field is normal to the interface So, the normal components of the fields are:

E_1n = 1 and E_2n = 2

From boundary condition we have

ε₁ E_1n – ε₂ E_2n = ρ_s

(where ρ_s is surface charge density on the interface)

(ε₀)(1) – (2ε₀)(2) = ρ_s

ρ_s = -3ε₀     

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

The decreasing current induces varying magnetic field around the wire as shown in above figure (Fleming’s left hand rule)

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:

$\mathrm{\nabla }\cdot E=\frac{{\rho }_v}{\epsilon }\ne 0$

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:

$\mathrm{\nabla }\times E=-\frac{\partial B}{\partial t}\ne 0$

$\mathrm{\nabla }.\cdot E=\frac{{\rho }_v}{\epsilon }\ne 0$

(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

$f=\frac{\omega }{2\pi }=\frac{10}{2\times \epsilon \times 2\times 8.854\times {10}^{-12}}$

Concept:

The normal component of the magnetic field satisfies the following boundary condition:

​​${\overrightarrow{B}}_{1n}={\overrightarrow{B}}_{2n}$

With B = μ H, the above expression can be written as:

${\mu }_1{\overrightarrow{H}}_{1n}={\mu }_2{\overrightarrow{H}}_{2n}$

And the tangential component of the magnetic field satisfies the following boundary condition:

H_1k – H_2k = k̅

$\frac{B_1}{{\mu }_1}-\frac{B_2}{{\mu }_2}=\overrightarrow{k}$

If the boundary is free of the current, i.e. if k̅ = 0:

${\overrightarrow{H}}_{1k}={\overrightarrow{H}}_{2k}$

$\frac{{\overrightarrow{B}}_{1k}}{{\mu }_1}=\frac{{\overrightarrow{B}}_{2k}}{{\mu }_2}$

Calculation:

From the boundary conditions, the normal component of flux density is uniform at the boundary.

i.e. B_1n = B_2n

B₁ sin θ₁ = B₂ sin θ₂

And the tangential component of field intensity is uniform for a current free boundary, i.e.

i.e. H_1t = H_2t

H₁ cos θ₁ = H₂ cos θ₂   

Derivation:

Consider the interface two different materials with dissimilar permeability μ1 and μ2.

Here $B_{t_1}$= tangential component on medium 1

$B_{N_1}$ = normal component on medium 1

$B_{t_1}$ = tangential component on medium

$B_{N_2}$ = Normal component on medium 2

Applying equation 1 on pull box and allowing Δh → 0

B1n ΔS – B2n ΔS = 0

${\overrightarrow{B}}_{1n}={\overrightarrow{B}}_{2n}$          (B = μH )

${\mu }_1{\overrightarrow{H}}_{1n}={\mu }_2{\overrightarrow{H}}_{2n}$

The normal component of $\overrightarrow{B}$ is continuous at the boundary.

The normal component of $\overrightarrow{H}$ is discontinuous at the boundary.

Tangential magnetic field:

Similarly, we can apply equation (2) to the enclosed both abcd shown in the figure below:

$k\mathrm{\Delta }\omega =H_{1}k\mathrm{\Delta }\omega +H_{1n}\frac{\mathrm{\Delta }\mathrm{h}}{2}+H_{2n}\cdot \frac{\mathrm{\Delta }h}{2}$

As Δh → 0

H1k – H2k = k

$\frac{B_1}{{\mu }_1}-\frac{B_2}{{\mu }_2}=k$    (B = μH)

Concept:

The circulation of the electric field around a loop is nothing but the electric potential, the magnitude of which is given by:

$V=\underset{C}{\oint }E\cdot dl$   ---(1)

Also, the potential generated around a loop is given by:

$V=-\frac{d\mathrm{\Phi }}{dt}$   ---(2)

From Equation (1) and (2), we get:

$\underset{C}{\oint }E\cdot dl=-\frac{d\mathrm{\Phi }}{dt}$

Calculation:

Given $\overrightarrow{B}=-\frac{2}{x}{a}_z$

The total magnetic flux passing through the loop will be:

$\mathrm{\Phi }=\int B\cdot dS$

$\varphi =\underset{x}{\overset{x+2}{\int }}\underset{y}{\overset{y+2}{\int }}\left(-\frac{2}{x}{a}_z\right)\cdot \left(-dxdy{a}_z\right)$

$=\left(2\ln \frac{x+2}{x}\right)\left(2\right)$

$=4\ln \left(\frac{x+2}{x}\right)$

Therefore, the circulation of the induced electric field in the loop will be:

$\underset{C}{\oint }E\cdot dl=-\frac{d\mathrm{\Phi }}{dt}$

$=-\frac{d}{dt}\left[4\ln \left(\frac{x+2}{x}\right)\right]$

$=-\frac{4}{\left(\frac{x+2}{x}\right)}\frac{d}{dt}\left(\frac{x+2}{x}\right)$

$=-\frac{4x}{x+2}\left(-\frac{2}{x^2}\frac{dx}{dt}\right)$

Since $\frac{dx}{dt}=v=2a_x$, the above expression becomes:

$=\frac{8}{x\left(x+2\right)}\left(2\right)$

$\underset{C}{\oint }E\cdot dl=\frac{16}{x\left(x+2\right)}$

The electric flux density is given by:

$\overrightarrow{D}=ϵ\overrightarrow{E}$

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

$D=\frac{ϵV}{d}$

Also, the displacement current density is defined as:

${\overrightarrow{J}}_d=\frac{\partial \overrightarrow{D}}{\partial t}=\frac{ϵ}{d}\frac{\partial V}{\partial t}$

$I_d=J_d.A=\frac{ϵA}{d}\frac{dV}{\partial t}$

Calculation:

Putting on the respective values, we get:

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

$=\frac{100}{\pi }\cos {10}^{3}tnA$

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

$E_{2n}=\frac{{\epsilon }_1}{{\epsilon }_2}{E}_{1n}=6a_z$

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