Question

A plane of electromagnetic waves travels in a medium with a relative permeability of 1.61 and relative permittivity of 6.44. If the magnitude of magnetic intensity is 4.5 × 10–2 Am–1 at a point, what will be the approximate magnitude of electric field intensity?(Given: Permeability of free space μ0 = 4π × 10–7 NA–2, speed of light in vacuum c = 3 × 108 ms–1)

• A. 16.96 Vm–1
• B. 2.25 × 10–2 Vm–1
• C. 8.48 Vm–1
• D. 6.75 × 106 Vm–1

Answer 1

Answer: (C)

8.48 Vm–1

Answer 2

H = 4.5 × 10–2
So B = μ0μ H
Thus
E=$\frac{c}{n}$B
(where n ⇒ refractive index)
So
E = $\frac{3\times10^8\times4\pi\times10{-7}\times1.61\times4.5\times10^{-2}}{\sqrt{1.61\times6.44}}$
E = 8.48
The correct option is (A): 16.96 Vm–1