Easy MCQ +4 / -1 PYQ · JEE Mains 2025

The unit of $\sqrt{\frac{2I}{\varepsilon_0 c}}$ is :

(I = intensity of an electromagnetic wave, c = speed of light)

  1. A <p>Vm</p>
  2. B <p>NC<sup>-1</sup></p> Correct answer
  3. C <p>NC</p>
  4. D <p>Nm</p>

Solution

<p>The unit of the expression $\sqrt{\frac{2I}{\varepsilon_0 c}}$ can be determined by exploring the relationship between intensity ($I$) and the electric field ($E_0$) in an electromagnetic wave.</p> <p>The intensity of an electromagnetic wave is given by:</p> <p>$ I = \frac{1}{2} \varepsilon_0 E_0^2 \times c $</p> <p>From this equation, we can solve for the electric field $E_0$:</p> <p>$ E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}} $</p> <p>The symbol $E_0$ represents the electric field, which is measured in units of Newtons per Coulomb (N/C). Therefore, the unit of the expression $\sqrt{\frac{2I}{\varepsilon_0 c}}$ is also $N/C$.</p> <p>Thus, the expression $\sqrt{\frac{2I}{\varepsilon_0 c}}$ has units of $N/C$.</p>

About this question

Subject: Physics · Chapter: Electromagnetic Waves · Topic: Properties of EM Waves

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