Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electromagnetic Waves

Question:

The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is:

Options:

c : 1

c2 : 1

1 : 1

$\sqrt{c}$ : 1

Correct Answer:

1 : 1

Explanation:

Intensity in terms of electric field, $u_{av}=\frac{1}{2} \varepsilon_0 E_0^2$

Intensity in terms of magnetic field, $u_{av}=\frac{1}{2} \frac{B_0^2}{\mu_0}$

Now, taking the intensity in terms of electric field.

$\left(U_{av}\right)_{\text {electric field }}=\frac{1}{2} \varepsilon_0 E_0^2$

$\Rightarrow=\frac{1}{2} \varepsilon_0\left(cB_0\right)^2$          $\left[∵ E_0=cB_0\right]$

$=\frac{1}{2} \varepsilon_0 \times c^2 B_0^2$

But, $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

∴ $\left(u_{av}\right)_{\text {electric field }}=\frac{1}{2} \varepsilon_0 \times \frac{1}{\mu_0 \varepsilon_0} B_0^2=\frac{1}{2} \frac{B_0^2}{\mu_0}$

= $\left(u_{av}\right)_{\text {magnetic field }}$

Thus, the energy in electromagnetic wave is divided equally between electric field vector and magnetic field vector. Therefore, the ratio of contributions by the electric field and magnetic field components to the intensity of an electromagnetic wave is 1 : 1.