Download PTU Transverse nature of EM Waves Question Bank

FirstRanker.com

Firstranker's choice www.FirstRanker.com

Maxwell's four equations are given as follows:

--- Content provided by FirstRanker.com ---

∇.D = ρ ...(1)

∇.B = 0 ...(2)

∇ × E = - ∂B/∂t ...(3)

∇ × H = J + ∂D/∂t ...(4)

However, D = ε₀E + P and B = μ₀(H + M) where D = ε₀E + P

--- Content provided by FirstRanker.com ---

J = σE

P = χₑε₀E

M = χₘH

For Vacuum, we have ρ = 0, J = 0

Maxwell's four equations simplify to the following form:

--- Content provided by FirstRanker.com ---

∇.E = 0 ...(5)

∇.B = 0 ...(6)

∇ × E = - ∂B/∂t ...(7)

∇ × B = (1/c²) ∂E/∂t ...(8)

Maxwell's differential form in terms of electric and magnetic fields can be written as follows:

--- Content provided by FirstRanker.com ---

E = E₀ e^(i(ωt - k.r)) ...(9)

B = B₀ e^(i(ωt - k.r)) ...(10)

Where,

r = xi + yj + zk = position vector

k = kx i + ky j + kz k

--- Content provided by FirstRanker.com ---

These (E, B) must simultaneously obey equations (5) to (8).

From equation (5), we can write:

kxEx + kyEy + kzEz = 0 ...(11)

Similarly, from equation (6), we can write:

kxBx + kyBy + kzBz = 0 ...(12)

--- Content provided by FirstRanker.com ---

Thus, electric field E ⊥ k (direction of propagation)

Similarly, we can show that B can satisfy equation (6) only if

B ⊥ k (electromagnetic wave)

To find relative orientation between E, B, k, we must validate equations (7) and (8) by the relations (9) and (10) of wave equation.

Now we can write:

--- Content provided by FirstRanker.com ---

∇ × E = (∂Ez/∂y - ∂Ey/∂z)i + (∂Ex/∂z - ∂Ez/∂x)j + (∂Ey/∂x - ∂Ex/∂y)k

Now (∇ × E)x = ∂Ez/∂y - ∂Ey/∂z

From equation (9), we have:

∂Ez/∂y = i ky Ez

∂Ey/∂z = i kz Ey

--- Content provided by FirstRanker.com ---

(∇ × E)x = i(kyEz - kzEy)

Similarly, we can find:

(∇ × E)y = i(kzEx - kxEz)

(∇ × E)z = i(kxEy - kyEx)

FirstRanker.com

--- Content provided by FirstRanker.com ---

Put values from equations (7) and (13), we get:

i(kxEy - kyEx) = - ∂B/∂t

Again from (10), we can show that:

∂B/∂t = iωB

Put values from equations (14) and (15) in (7), we get:

--- Content provided by FirstRanker.com ---

i(k × E) = -iωB

or k × E = ωB

Equation (16) shows that B ⊥ k and E.

From equations (5) and (6), it is clear that E, B, k are mutually perpendicular to each other.

Now, we can also satisfy equation (8) as follows:

--- Content provided by FirstRanker.com ---

∇ × B = μ₀J + με₀ ∂E/∂t

However, at low temperature of a polarizable dielectric medium:

J = -σE

We can see that:

∇ × B = i(k × B)

--- Content provided by FirstRanker.com ---

∂E/∂t = iωE

Put these values in (8), we get:

i(k × B) = μ₀(-σE) + με₀(iωE)

The equation above satisfies Maxwell's equation.

Hence electromagnetic

--- Content provided by FirstRanker.com ---