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

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 ]

TestHub · Accepted Answer

The relation between magnetic field and electric field is given by,

$c = \frac{E_{0}}{B_{0}} \Rightarrow B_{0} = \frac{E_{0}}{c}$

And magnetic intensity is given by,

$H_{0} = \frac{B_{0}}{μ_{0}} \Rightarrow H_{0} = \frac{E_{0}}{c μ_{0}}$

Also, $\vec{E}$ and $\vec{B}$ are perpendicular to each other.

Therefore,

$|H_{y 0}| = \frac{301.6}{3 \times 10^{8} \times 4 π \times 10^{- 7}} = 0.8 A m^{- 1} |H_{x 0}| = \frac{452.4}{3 \times 10^{8} \times 4 π \times 10^{- 7}} = 1.2 A m^{- 1}$

Hence, C is the answer.

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