EM Waves

Q: If the magnetic field in a plane electromagnetic wave is given by $\vec{B} = 3 \times 10^{-8} sin(1.6 \times 10^3 x + 48 \times 10^{10} t)\hat{j}\; T $ , then what will be the expression for electric field ?

$(a) \vec{E} = 9 sin (1.6 \times 10^3 x + 48 \times 10^{10}t ) \hat{k} \; V/m$

$(b) \vec{E} = 60 sin (1.6 \times 10^3 x + 48 \times 10^{10}t ) \hat{k} \; V/m$

$(c) \vec{E} = 3 \times 10^{-8} sin (1.6 \times 10^3 x + 48 \times 10^{10}t ) \hat{i} \; V/m$

$(d) \vec{E} = 3 \times 10^{-8} sin (1.6 \times 10^3 x + 48 \times 10^{10}t ) \hat{j} \; V/m$

Q: A passenger in an aeroplane shall

(a) Never see a rainbow

(b) May see a primary and a secondary rainbow as concentric circles

(c) May see a primary and a secondary rainbow as concentric arcs

(d) Shall never see a secondary rainbow

Ans: (b)

Q: A plane EM wave travelling along z– direction is described by E = E₀ sin(kz – ωt)j ̂ . Show that

(i) The average energy density of the wave is given by $u_{av} = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} \frac{B_0^2}{\mu_0} $

(ii) The time averaged intensity of the wave is given by $ I_{av} = \frac{1}{2} c \epsilon_0 E_0^2 $

Q: Sea water at frequency v = 4 × 10⁸ Hz has permittivity ε ≈ 80 ε₀ , permeability μ ≈ μ₀ and resistivity ρ = 0.25 m . imagine a parallel plate capacitor immersed in sea water and driven by an alternation voltage source V(t) = V₀ sin (2πvt). What fraction of the conduction current density is the displacement current density ?

Sol: Suppose d =separation between plates of capacitor immersed in sea water & applied voltage across the plates  is V(t) = V₀ sin (2πvt)

Electric field between the plates , $E = \frac{V(t)}{d}$

$E = \frac{V_0 }{d}sin2\pi \nu t $

Conduction current density , $\large  J^c = \frac{E}{\rho} = \frac{V_0 }{\rho d}sin2\pi \nu t$

$\large  J^c  =  J_0^c sin2\pi \nu t$ ; Where $\large  J_0^c = \frac{V_0 }{\rho d}$

Displacement current density , $\large  J^d = \epsilon \frac{dE}{dt}$

$\large  J^d = \epsilon \frac{d}{dt}(\frac{V_0 }{d}sin2\pi \nu t) $

$\large  J^d = \frac{\epsilon 2\pi \nu V_0}{d} cos (2\pi \nu t ) $

$\large  J^d = J_0^d cos (2\pi \nu t ) $ ; Where $\large  J_0^d = \frac{\epsilon 2\pi \nu V_0}{d}$

$\large \frac{J_0^d }{J_0^c} = \frac{\frac{\epsilon 2\pi \nu V_0}{d}}{\frac{V_0 }{\rho d}} $

$\large \frac{J_0^d }{J_0^c} = 2 \pi \nu \epsilon \rho$

$\large \frac{J_0^d }{J_0^c} = 2 \pi \nu (80\epsilon_0 ) \rho$

$ = 2 \pi \nu (80\epsilon_0 ) \times 0.25 = 4 \pi \epsilon_0 \nu \times 10 $

$= \frac{4 \times 10^8 \times 10}{9 \times 10^9} = \frac{4}{9}$