If electric field intensity of a uniform plane electromagnetic wave is given asE=-301.6 sin(kz-ωt)a^x+452.4 sin(kz-ωt)a^yVmThen, magnetic intensity 'H' of this wave in Am-1 will be : [Given: Speed of light in vacuum c=3×108 ms-1, permeability of vacuum μ0=4π×10-7NA-2]

- 0 . 8 sin ( k z - ω t ) a ^ y - 1 . 2 sin ( 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 ( k z - ω t ) a ^ x

+ 0 . 8 sin ( k z - ω t ) a ^ y + 0 . 8 sin ( k z - cot ) 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 - cot ) a ^ z

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

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$+ 1.0 \times 10^{- 6} \sin (k z - ω t) {\hat{a}}_{y} + 1.5 \times 10^{- 6} (k z - ω t) {\hat{a}}_{x}$

$+ 0.8 \sin (k z - ω t) {\hat{a}}_{y} + 0.8 \sin (k z - \cot) {\hat{a}}_{x}$

$- 0.8 \sin (k z - ω t) {\hat{a}}_{y} - 1.2 \sin (k z - ω t) {\hat{a}}_{x}$

$- 1.0 \times 10^{- 6} \sin (k z - ω t) {\hat{a}}_{y} - 1.5 \times 10^{- 6} \sin (k z - \cot) {\hat{a}}_{z}$

answer is C.

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Phase of electric field is $(k z - ω t)$ . This means direction of wave propagation is +z axis.

To find direction of $\vec{B}$

$\vec{C} \times \vec{E} \to gives direction of \vec{B}$

$For first part \hat{E} = - \hat{i}, \Rightarrow \hat{B} = - \hat{j}$

$For second part \hat{E} = \hat{j}, \Rightarrow \hat{B} = - \hat{i}$

Using, $\vec{H} = \frac{\vec{B}}{μ_{0}} = \frac{E}{μ_{0} c}$

$\vec{H} = - 0.8 \sin (kz - ωt) {\hat{a}}_{y} - 1.2 \sin (kz - ωt) {\hat{a}}_{x}$

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