JEE Main 2019 (Online) 10th January Morning Slot | Electromagnetic Waves Question 121 | Physics

If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 10⁸ B = 100 × 10^–6 sin $$\left[ {2\pi \times 2 \times {{10}^{15}}\left( {t - {x \over c}} \right)} \right]$$ then the maximum electric field associated with it is -

4.5 $$ \times $$ 10⁴ N/C

4 $$ \times $$ 10⁴ N/C

6 $$ \times $$ 10⁴ N/C

3 $$ \times $$ 10⁴ N/C

JEE Main 2019 (Online) 9th January Evening Slot

MCQ (Single Correct Answer)

-1

The energy associated with electric field is (U_E) and with magnetic field is (U_B) for an electromagnetic wave in free space. Then :

$${U_E} = {{{U_B}} \over 2}$$

$${U_E} > {U_B}$$

$${U_E} < {U_B}$$

$${U_E} = {U_B}$$

JEE Main 2019 (Online) 9th January Morning Slot

MCQ (Single Correct Answer)

-1

A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, $$\overrightarrow E = 6.3\widehat j\,V/m.$$ The corresponding magnetic field $$\overrightarrow {B,} $$ at that point will be :

18.9 $$ \times $$ 10^$$-$$8 $$\widehat k$$T

2.1 $$ \times $$ 10^$$-$$8 $$\widehat k$$T

6.3 $$ \times $$ 10^$$-$$8 $$\widehat k$$T

18.9 $$ \times $$ 10⁸ $$\widehat k$$T

JEE Main 2018 (Online) 16th April Morning Slot

MCQ (Single Correct Answer)

-1

A plane electromagnetic wave of wavelength $$\lambda $$ has an intensity I. It is propagatting along the positive Y-direction. The allowed expressions for the electric and magnetic fields are givn by :