1
JEE Main 2024 (Online) 1st February Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
If frequency of electromagnetic wave is $60 \mathrm{~MHz}$ and it travels in air along $z$ direction then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other and the wavelength of the wave (in $\mathrm{m}$ ) is :
A
2.5
B
5
C
10
D
2
2
JEE Main 2024 (Online) 31st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Given below are two statements:

Statement I: Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields.

Statement II: When electromagnetic waves strike a surface, a pressure is exerted on the surface.

In the light of the above statements, choose the most appropriate answer from the options given below:

A
Statement I is incorrect but Statement II is correct.
B
Both Statement I and Statement II are correct.
C
Statement I is correct but Statement II is incorrect.
D
Both Statement I and Statement II are incorrect.
3
JEE Main 2024 (Online) 31st January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of $$5 \times 10^{10} \mathrm{~Hz}$$ and an amplitude of $$50 \mathrm{~Vm}^{-1}$$. The total average energy density of the electromagnetic field of the wave is : [Use $$\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2$$ ]

A
$$4.425 \times 10^{-8} \mathrm{Jm}^{-3}$$
B
$$2.212 \times 10^{-10} \mathrm{Jm}^{-3}$$
C
$$2.212 \times 10^{-8} \mathrm{Jm}^{-3}$$
D
$$1.106 \times 10^{-8} \mathrm{Jm}^{-3}$$
4
JEE Main 2024 (Online) 30th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The electric field of an electromagnetic wave in free space is represented as $$\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{i}$$. The corresponding magnetic induction vector will be :

A
$$\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}+\mathrm{k} z) \hat{j}$$
B
$$\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{j}$$
C
$$\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}-\mathrm{k} z) \hat{j}$$
D
$$\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}+\mathrm{kz}) \hat{j}$$
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