Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The ratio of the magnitude of the magnetic field and electric field intensity of a plane electromagnetic wave in free space of permeability $\mu_0$ and permittivity $\varepsilon_0$ is: (Given that $c=$ velocity of light in free space)

$c$

$\frac{1}{c}$

$\frac{c}{\sqrt{\mu_0\varepsilon_0}}$

$\frac{\sqrt{\mu_0\varepsilon_0}}{c}$

Step-by-Step Solution

Relationship between E and B: In a plane electromagnetic wave propagating in free space, the ratio of the magnitude of the electric field ( $E_0$ ) to the magnitude of the magnetic field ( $B_0$ ) is equal to the speed of light ( $c$ ). $c = \frac{E_0}{B_0}$ (Reference: NCERT Class 12, Physics Part I, Chapter 8, Section 8.3).
Ratio Asked: The question asks for the ratio of the magnetic field to the electric field ( $\frac{B_0}{E_0}$ ).
Calculation: Rearranging the standard formula: $\frac{B_0}{E_0} = \frac{1}{c}$
Additional Context: Since the speed of light is also defined as $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ , the ratio could also be expressed as $\sqrt{\mu_0 \varepsilon_0}$ , but $\frac{1}{c}$ is the option provided.

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