The electric field for an electromangetic wave in free space is E=i 30 (k z-5 × 10^8 t) , where magnitude of E is in V / m . The magnitude of wave vector, k is (velocity of em wave in free space =3 × 10^8 ~m / s )

Correct answer: C. 1.66 rad ^ -1 The given E=i 30 (k z-5 ×10^8 t) We know that, E= iV (k z- ) Comparing the both equations, ω=5 ×10^8 rad / s But we know that also K=(2 π)/(λ) and ω= 2 π where, λ is wavelength and v is frequency of the wave array ll & (ω)/(k)=(2 π v)/(2 π / λ)=v λ=C & C=(ω)/(k) or & k=(ω)/(C)=(5 × 10^8)/(3 × 10^8) & k=(5)/(3)=1.66 rad/ m array

JEE Main

Exam: JEE Main
Subject: Physics
Chapter: Electromagnetic Waves

JEE Main · Physics · Electromagnetic Waves

The electric field for an electromangetic wave in free space is E=i...

The electric field for an electromangetic wave in free space is

\mathbf{E}=\mathbf{i} 30 \cos \left(k z-5 \times 10^8 t\right)

, where magnitude of

E

is in

\mathrm{V} / \mathrm{m}

. The magnitude of wave vector,

k

is (velocity of em wave in free space

=3 \times 10^8 \mathrm{~m} / \mathrm{s} )

Options

A $0.46 \mathrm{rad} \mathrmm^{-1}$
B $3 \mathrm{rad} \mathrm{m}{ }^{-1}$
C $1.66 \mathrm{rad} \mathrmm^{-1}$
D $0.83 \mathrm{rad} \mathrm{m} \mathrmm^{-1}$

Correct answer

C. $1.66 \mathrm{rad} \mathrmm^{-1}$

Step-by-step solution

The given

\mathbf{E}=\mathbf{i} 30 \cos \left(k z-5 \times10^8 t\right)

We know that,

\mathbf{E}= \mathrm{iV} \cos (k z-\omegat)

Comparing the both equations,

\omega=5 \times10^8

\mathrm{rad}

/ \mathrm{s}

But we know that also
K=\frac{2 \pi}{\lambda} \text { and } \omega=

2 \pi \nu

where,

\lambda

is wavelength and

v

is frequency of the wave

\begin{array}{ll}<br/>\therefore & \frac{\omega}{k}=\frac{2 \pi v}{2 \pi / \lambda}=v \lambda=C \\<br/>\Rightarrow & C=\frac{\omega}{k}\\<br/>\text {or }& k=\frac{\omega}{C}=\frac{5 \times 10^8}{3 \times 10^8}\\<br/>\Rightarrow & k=\frac{5}{3}=1.66 \mathrm{rad}/ \mathrm{m}<br/>\end{array}

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