If Electric field intensity of a uniform plane electro magnetic wave is given as E = - 301 . 6 sin k z - ω t a ^ x + 452 . 4 sin k z - ω t a ^ y V m - 1 . Then, magnetic intensity H of this wave in A m - 1 will be [Given : Speed of light in vacuum c = 3 × 10 8 m s - 1 , Permeability of vacuum μ 0 = 4 π × 10 - 7 N A - 2 ]

Question

If Electric field intensity of a uniform plane electro magnetic wave is given as E = - 301 . 6 sin k z - ω t a ^ x + 452 . 4 sin k z - ω t a ^ y V m - 1 . Then, magnetic intensity H of this wave in A m - 1 will be [Given : Speed of light in vacuum c = 3 × 10 8 m s - 1 , Permeability of vacuum μ 0 = 4 π × 10 - 7 N A - 2 ], + 0 . 8 sin k z - ω t a ^ y + 0 . 8 sin k z - ω t a ^ x, + 1 . 0 × 10 - 6 sin k z - ω t a ^ y + 1 . 5 × 10 - 6 k z - ω t a ^ x, - 0 . 8 sin k z - ω t a ^ y - 1 . 2 sin k z - ω t a ^ x, - 1 . 0 × 10 - 6 sin k z - ω t a ^ y - 1 . 5 × 10 - 6 sin k z - ω t a ^ x

MARKS App · Accepted Answer

The relation between magnetic field and electric field is given by, c = E 0 B 0 ⇒ B 0 = E 0 c And magnetic intensity is given by, H 0 = B 0 μ 0 ⇒ H 0 = E 0 c μ 0 Also, E → and B → are perpendicular to each other. Therefore, H y 0 = 301 . 6 3 × 10 8 × 4 π × 10 - 7 = 0 . 8 A m - 1 H x 0 = 452 . 4 3 × 10 8 × 4 π × 10 - 7 = 1 . 2 A m - 1 Hence, C is the answer.

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