If electric field intensity of a uniform plane electro magnetic wave is given as $\mathrm{E}=-301.6 \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{x}}+452.4 \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{a}_{y} \frac{V}{m}$ Then, magnetic intensity H of this wave in $\mathrm{Am}^{-1}$ will be: [Given: Speed of light in vacuum c $=3 \times 10^{8} \mathrm{~ms}^{-1}$, permeability of vacuum $\mu_{0}=4 \pi \times 10^{-7} \mathrm{NA}^{-2}$ ]

(C) $-0.8 \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{y}}-1.2 \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{x}}$

(A) $+0.8 \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{y}}+0.8 \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{x}}$

(B) $+1.0 \times 10^{-6} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{y}}+1.5 \times 10^{-6}(\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{x}}$

(D) $-1.0 \times 10^{-6} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{y}}-1.5 \times 10^{-6} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{x}}$

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · EM Waves

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If electric field intensity of a uniform plane electro magnetic wave is given as

E = - 301.6 sin (kz - ω t) \overset{a}{^}_{x} + 452.4 sin (kz - ω t) \overset{a}{^}_{y} \frac{V}{m}

Then, magnetic intensity H of this wave in

Am^{- 1}

will be:
[Given: Speed of light in vacuum c

= 3 \times 1 0^{8} ms^{- 1}

, permeability of vacuum

μ_{0} = 4 π \times 1 0^{- 7} NA^{- 2}

]

(A) $+ 0.8 sin (kz - ω t) \overset{a}{^}_{y} + 0.8 sin (kz - ω t) \overset{a}{^}_{x}$
(B) $+ 1.0 \times 1 0^{- 6} sin (kz - ω t) \overset{a}{^}_{y} + 1.5 \times 1 0^{- 6} (kz - ω t) \overset{a}{^}_{x}$
(C) $- 0.8 sin (kz - ω t) \overset{a}{^}_{y} - 1.2 sin (kz - ω t) \overset{a}{^}_{x}$ ✓ Correct
(D) $- 1.0 \times 1 0^{- 6} sin (kz - ω t) \overset{a}{^}_{y} - 1.5 \times 1 0^{- 6} sin (kz - ω t) \overset{a}{^}_{x}$

Answer: (C)

Solution

For an EM wave,

E \times B = c \hat{k}

(direction of propagation). The magnetic field amplitude is

B = E / c

and

H = B / μ_{0} = E / (μ_{0} c)

. For the

- \overset{a}{^}_{x}

component of

E

\hat{E} \times \hat{B} = \hat{k}

gives

\hat{B} = - \overset{a}{^}_{y}

, so

H_{y} = - 301.6/ (μ_{0} c) = - 0.8

A/m. For the

+ \overset{a}{^}_{y}

component:

\hat{E} \times \hat{B} = \hat{k}

gives

\hat{B} = - \overset{a}{^}_{x}

, so

H_{x} = - 452.4/ (μ_{0} c) = - 1.2

A/m. Thus

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

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 26 Jun 2022 · Shift 1 · EM Waves

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