Mumbai University > Electronics Engineering > Sem 5 > Electromagnetic Engineering

Marks: 10 Marks

Year: May 2015

Maxwell’s equation for any medium is given as:

$∇.\bar{E}=0 –(1) \\ ∇.\bar{H}=0-(2) \\ ∇ ×\bar{E}= -jwμ \bar{H}- (3) \\ ∇ × \bar{H}=(σ+jwε) \bar{E}- (4)$

Consider 4th equation

$∇ × \bar{H }=(σ+jwε) \bar{E}$

Taking cross product on both sides

$∇ × ∇ × \bar{H}= (σ+jwε) ∇ × \bar{E} \\ ∇ (∇.\bar{H } )- ∇^2 \bar{H }=(σ+jwε)(-jwμ) \bar{H} \\ ∇^2 \bar{H}= (σ+jwε)(jwμ) \bar{H} $

Similarly,

$∇ ×\bar{E}= -jwμ\bar{H}$

Taking cross product on both sides,

$∇ × ∇ × \bar{E}= (-jwμ) ∇ × \bar{H} \\ ∇ (∇.\bar{E} )- ∇^2 \bar{E}=(-jwμ)(σ+jwε) \bar{E } \\ ∇^2 \bar{E}= (σ+jwε)(jwμ) \bar{E}$

For free space (σ=0)

$∇^2 \bar{H}= -w^2 μ\bar{H}ε \\ ∇^2 \bar{E}= -w^2 με\bar{E}$