The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is \overrightarrow{E}=E_{0} \hat{j} \cos (\omega t-k x). The magnetic field \overrightarrow{B}, at the moment t=0 is: [3 Sep 2020 Shift 2]

Correct Answer is : →B=E0√µ0ε0cos(kx)∧k

\overrightarrow{B}= {E_{0}}/\sqrt{\mu_{0} \epsilon_{0}}} \cos (k x) \hat{k}

\overrightarrow{B}=E_{0} \sqrt{\mu_{0} \epsilon_{0}} \cos (k x) \hat{j}

\overrightarrow{B}=E_{0} \sqrt{\mu_{0} \epsilon_{0}} \cos (k x) \hat{k}

\overrightarrow{B}= {E_{0}}/\sqrt{\mu_{0} \epsilon_{0}}} \cos (k x) \hat{j}

JEE Mains 03-Sep-2020 Shift 2 Solved Paper

Section: Physics

Question : 12 of 75

Marks: +1, -0

The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is

→

=E0

∧

‌cos(ωt−kx). The magnetic field

→

, at the moment t=0 is:

[3 Sep 2020 Shift 2]

Go to Question:

Prev Question

Next Question