An electromagnetic wave propagates in a medium having a relative permittivity of $2$ . If the ratio of the magnitude of the electric field intensity to the magnetic field intensity for this wave is $240 \pi \, \Omega$ , the relative permeability of the medium is:
(Given $\sqrt{\dfrac{\mu_0}{\varepsilon_0}} = 120 \pi \, \Omega$ )

Question

An electromagnetic wave propagates in a medium having a relative permittivity of

2

. If the ratio of the magnitude of the electric field intensity to the magnetic field intensity for this wave is

240 \pi \, \Omega

, the relative permeability of the medium is:
(Given

\sqrt{\dfrac{\mu_0}{\varepsilon_0}} = 120 \pi \, \Omega

)

Quantrex Academy · Accepted Answer

Correct answer: A. 8

The ratio of the magnitude of the electric field intensity (

E

) to the magnetic field intensity (

H

) represents the wave impedance (

Z

) of the medium.

The wave impedance of a medium is given by:

\dfrac{E}{H} = \sqrt{\dfrac{\mu}{\varepsilon}} = \sqrt{\dfrac{\mu_0 \mu_r}{\varepsilon_0 \varepsilon_r}}

We can rewrite this as:

\dfrac{E}{H} = \sqrt{\dfrac{\mu_0}{\varepsilon_0}} \times \sqrt{\dfrac{\mu_r}{\varepsilon_r}}

Substitute the given values into the equation:

240 \pi = 120 \pi \times \sqrt{\dfrac{\mu_r}{2}}

Dividing both sides by

120 \pi

gives:

2 = \sqrt{\dfrac{\mu_r}{2}}

Squaring both sides:

4 = \dfrac{\mu_r}{2}

\mu_r = 8

The relative permeability of the medium is

8

.

Answer:

8

An electromagnetic wave propagates in a medium having a relative...