A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 × 10^-2 Am^-1 at a point, what will be the approximate magnitude of electric field intensity at that point ?

(Given : Permeability of free space µ₀ = 4p × 10^-⁷ NA^-², speed of light in vacuum c = 3 × 10⁸ ms^-¹)

Option 1 -

16.96 Vm^-1

Option 2 -

16.96 Vm^-1

Option 3 -

16.96 Vm^-1

Option 4 -

16.96 Vm^-1

4 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

Velocity of wave in a medium = $\frac{1}{\sqrt[]{μ_{m} \in_{m}}}$ $\frac{}{_{}_{}}$

$\begin{array}{l} v = \frac{1}{\sqrt[]{μ_{0} μ_{r} \in_{r} \in_{0}}} \\ v = \frac{3 \times 1 0^{8}}{\sqrt[]{1.61 \times 6.44}} = 0.931 \times 1 0^{8} m / s e c \end{array}$

$\begin{array}{l} B = μ_{m} H \\ = μ_{r} μ_{0} H \\ = 1.61 \times 4 π \times 1 0^{- 7} \times 4.5 \times 1 0^{- 2} \\ = 1.61 \times 4 π \times 4.5 \times 1 0^{- 9} \end{array}$

$\frac{E}{B} = v$

E = vB = 0.931 × 10⁸ × 1.61 × 4p × 4.5 × 10^-⁹

= [0.931 × 1.61 × 4p × 4.5] × 10^-¹

= 8.476

» 8.476 v m^-¹

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