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 - <p>16.96 Vm<sup>-1</sup></p>
Option 2 - <p>16.96 Vm<sup>-1</sup></p>
Option 3 - <p>16.96 Vm<sup>-1</sup></p>
Option 4 - <p>16.96 Vm<sup>-1</sup></p>
5 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
V
6 months ago
Correct Option - 3
Detailed Solution:

Velocity of wave in a medium = 1 μ m m

v = 1 μ 0 μ r r 0 v = 3 * 1 0 8 1 . 6 1 * 6 . 4 4 = 0 . 9 3 1 * 1 0 8 m / s e c

B = μ m H = μ r μ 0 H = 1 . 6 1 * 4 π * 1 0 7 * 4 . 5 * 1 0 2 = 1 . 6 1 * 4 π * 4 . 5 * 1 0 9

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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Physics Ncert Solutions Class 12th 2023

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