Question

JEE MAINS PHYSICS QUESTION SEPTEMBER 2020

Answer 1

Answer:

$\boxed{\mathfrak{(1) \ 9sin(ky + \omega t)( \hat{-k}) V/m \ and \ (2) \ 9sin(ky + \omega t)( \hat{k}) \ V/m}}$

Given:

$\sf Magnetic \: field \: (\overrightarrow{B}) = 3 \times {10}^{ - 8} sin(ky + \omega t)( \hat{i})$

To Find:

$\sf Electric \: field \: ( \overrightarrow{E})$

Explanation:

$\bigstar$ In electromagnetic wave Electric field ($\sf \overrightarrow{E}$) & Magnetic field ($\sf \overrightarrow{B}$) are perpendicular to each other.

$\bigstar$ In electromagnetic wave Electric field and Magnetic field are related to each other as:

$\boxed{ \bold{c = \frac{\overrightarrow{E}}{\overrightarrow{B}}}}$

Where:

c $\rightarrow$ speed of light

So,

$\sf \implies \overrightarrow{E} = \overrightarrow{B}c \\ \\ \sf \implies \overrightarrow{E} = 3 \times {10}^{ - 8} sin(ky + \omega t) \times 3 \times {10}^{8} ( \hat{k} \: or \: \hat{ -k }) \\ \\ \sf \implies \overrightarrow{E} = 9sin(ky + \omega t)( \hat{k} \: or \: \hat{ -k })$