JEE Main 2025 (Online) 24th January Evening Shift | Electromagnetic Waves Question 13 | Physics

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JEE Main 2025 (Online) 24th January Evening Shift

MCQ (Single Correct Answer)

-1

Arrange the following in the ascending order of wavelength $(\lambda)$ :

(A) Microwaves $\left(\lambda_1\right)$

(B) Ultraviolet rays $\left(\lambda_2\right)$

(D) X-rays $\left(\lambda_4\right)$

Choose the most appropriate answer from the options given below :

$\lambda_4<\lambda_3<\lambda_2<\lambda_1$

$\lambda_4<\lambda_2<\lambda_3<\lambda_1$

$\lambda_3<\lambda_4<\lambda_2<\lambda_1$

$\lambda_4<\lambda_3<\lambda_1<\lambda_2$

JEE Main 2025 (Online) 23rd January Evening Shift

MCQ (Single Correct Answer)

-1

A plane electromagnetic wave of frequency 20 MHz travels in free space along the $+x$ direction. At a particular point in space and time, the electric field vector of the wave is $\mathrm{E}_y=9.3 \mathrm{Vm}^{-1}$. Then, the magnetic field vector of the wave at that point is

$\mathrm{B}_z=1.55 \times 10^{-8} \mathrm{~T}$

$\mathrm{B}_z=6.2 \times 10^{-8} \mathrm{~T}$

$\mathrm{B}_z=3.1 \times 10^{-8} \mathrm{~T}$

$\mathrm{B}_z=9.3 \times 10^{-8} \mathrm{~T}$

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 :

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

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

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

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

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