If electric field intensity of uniform plane electro magnetic wave is given asE-301-6sinkz- t-x-452-4sinkz- t-yVmThen-magnetic intensity -H- of this wave in Am-1 will be-Given-Speed of light in vacuum c-3- 108ms-1-Permiability of vacuum -0-4- 10-7NA-2-

We know nbsp; B×C=E Taking cross product of C both the sides C× (B×C)=C×E So nbsp; nbsp;B=C×EC^2 nbsp;C= C k̂ E= 301.6sin(kz ω t)â_x+452.4sin(kz ...

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Byju's Answer

Standard XII

Physics

Properties of Field Lines

If electric f...

Question

If electric field intensity of uniform plane electro magnetic wave is given as
$E = - 301.6 sin (k z - ω t) {^a}_{x} + 452.4 sin (k z - ω t) {^a}_{y} \frac{V}{m}$
Then,magnetic intensity $^{'} H^{'}$ of this wave in $A m^{- 1}$ will be:
[Given:Speed of light in vacuum $c = 3 \times 10^{8} m s^{- 1}$ ,Permiability of vacuum $μ_{0} = 4 π \times 10^{- 7} N A^{- 2}$ ]

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Solution

We know
$\to B \times \to C = \to E$
Taking cross product of $\to C$ both the sides

$\to C \times (\to B \times \to C) = \to C \times \to E$

So $\to B = \frac{\to C \times \to E}{C^{2}}$

$\to C = C^k$

$\to E = - 301.6 sin (k z - ω t) {^a}_{x} + 452.4 sin (k z - ω t) {^a}_{y}$

$and \to H = \frac{\to B}{μ_{0}}$

On solving

$\to H = - 0.8 sin (k z - ω t) {^a}_{y} - 1.2 sin (k z - ω t) {^a}_{x}$

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If electric field intensity of uniform plane electro magnetic wave is given as E=−301.6sin(kz−ωt)^ax+452.4sin(kz−ωt)^ayVm Then,magnetic intensity ′H′ of this wave in Am−1 will be: [Given:Speed of light in vacuum c=3×108ms−1,Permiability of vacuum μ0=4π×10−7NA−2]

We know →B×→C=→E Taking cross product of →C both the sides →C×(→B×→C)=→C×→E So →B=→C×→EC2 →C=C ^k →E=−301.6sin(kz−ωt)^ax+452.4sin(kz−ωt)^ay and →H=→Bμ0 On solving →H=−0.8sin(kz−ωt)^ay−1.2sin(kz−ωt)^ax