JEE Main 2025 (Online) 23rd January Morning Shift | Electromagnetic Waves Question 16 | Physics

JEE Main 2025 (Online) 23rd January Morning Shift

MCQ (Single Correct Answer)

-1

The electric field of an electromagnetic wave in free space is $\overrightarrow{\mathrm{E}}=57 \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](4 \hat{i}-3 \hat{j}) N / C$. The associated magnetic field in Tesla is

$\overrightarrow{\mathrm{B}}=\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](\hat{k})$

$\overrightarrow{\mathrm{B}}=\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})$

$\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](\hat{k})$

$\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})$

JEE Main 2024 (Online) 9th April Evening Shift

MCQ (Single Correct Answer)

-1

The magnetic field in a plane electromagnetic wave is $$\mathrm{B}_{\mathrm{y}}=\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{T}$$. The corresponding electric field will be :

$$E_z=105 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}$$

$$E_y=10.5 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}$$

$$E_y=1.17 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}$$

$$E_z=1.17 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}$$

JEE Main 2024 (Online) 9th April Morning Shift

MCQ (Single Correct Answer)

-1

A plane EM wave is propagating along $$x$$ direction. It has a wavelength of $$4 \mathrm{~mm}$$. If electric field is in $$y$$ direction with the maximum magnitude of $$60 \mathrm{~Vm}^{-1}$$, the equation for magnetic field is :

$$\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}$$

$$\mathrm{B}_z=2 \times 10^{-7} \sin \left[\frac{\pi}{2} \times 10^3\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}$$

$$\mathrm{B}_z=60 \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}$$

$$\mathrm{B}_x=60 \sin \left[\frac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{i}} \mathrm{T}$$

JEE Main 2024 (Online) 8th April Morning Shift

MCQ (Single Correct Answer)

-1

Average force exerted on a non-reflecting surface at normal incidence is $$2.4 \times 10^{-4} \mathrm{~N}$$. If $$360 \mathrm{~W} / \mathrm{cm}^2$$ is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:

$$20 \mathrm{~m}^2$$

$$0.2 \mathrm{~m}^2$$

$$0.1 \mathrm{~m}^2$$

$$0.02 \mathrm{~m}^2$$

PREVIOUS NEXT

JEE Main Subjects