Physics MCQs for NEET — Practice Questions with Answers

The relationship between the electric field (E) and the magnetic field (B) in an electromagnetic wave is given by:

$$ B = rac{E}{c} $$

Given, E = $900 ext{ NC}^{-1}$ and c = $3 imes 10^8 ext{ m/s}$, we can calculate the magnetic field (B) as:

$$ B = rac{900}{3 imes 10^8} = 3 imes 10^{-6} ext{ T} $$

Hence, the correct option is $3 imes 10^{-6} ext{ T}$.

Electromagnetic waves are produced by an accelerating charge. When a charge accelerates, it creates a changing electric field, which in turn creates a changing magnetic field, leading to the production of electromagnetic waves. This is a fundamental concept in electromagnetism.

Maxwell's equations are derived from the laws of both electricity and magnetism. These four equations describe how electric fields and magnetic fields interact and propagate. They are crucial for understanding classical electromagnetism.

Radio waves have the longest wavelength among the given options. The electromagnetic spectrum is ordered by wavelength, with radio waves having the longest wavelengths, followed by infrared, visible light, and then X-rays with the shortest wavelengths.

Gamma rays (γ rays) have the highest frequency among the electromagnetic waves listed. The electromagnetic spectrum is ordered by increasing frequency as follows: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Hence, γ rays have the highest frequency.

The maximum value of the magnetic field (B) in an electromagnetic wave can be calculated using the relation between the electric field (E) and the magnetic field (B) given by the speed of light (c): E = cB. Given that the speed of light, c = 3.0 × 10^8 m/s and E = 6.0 × 10^-4 V/m, we can solve for B:

B = E / c = (6.0 × 10^-4 V/m) / (3.0 × 10^8 m/s) = 2.0 × 10^-12 T.

Intensity of radiation is the power per unit area. The dimensional formula for power is $[M^1 L^2 T^{-3}]$ and for area is $[L^2]$. Therefore, the dimensional formula for intensity is obtained by dividing the dimensional formula of power by that of area, which gives $[M^1 L^0 T^{-3}]$.

The frequency of an electromagnetic wave does not change when it passes from one medium to another. It remains the same. Therefore, the frequency remains 3 MHz even when the wave passes through a medium with relative permeability $\varepsilon_r = 4.0$.

The frequency of an electromagnetic wave is related to its wavelength and the speed of light by the formula $\nu = \frac{c}{\lambda}$, where $\nu$ is the frequency, $c$ is the speed of light ($3 \times 10^8$ m/s), and $\lambda$ is the wavelength. For a wavelength of 25 mm (or $25 \times 10^{-3}$ m), the frequency is $\nu = \frac{3 \times 10^8}{25 \times 10^{-3}} = 1.2 \times 10^{10}$ Hz.