Question 705035

SCQMEDIUM

The electric field of an electromagnetic wave in free space is given by $\vec{E} = E_0 \sin(\omega t - k_z z)\hat{j}$ . The magnetic field $\vec{B}$ vanishes for $t = \frac{k z z}{\omega}$ . The Poynting vector of the system is

(A)

$\frac{k_z}{2 \mu_0 \omega} E_0^2 \sin^2(\omega t - k_z z) \hat{k}$

(B)

$\frac{4 k_z}{\mu_0 \omega} E_0^2 \sin^2(\omega t - k_z z) \hat{k}$

(C)

$\frac{2 k_z}{\mu_0 \omega} E_0^2 \sin^2(\omega t - k_z z) \hat{k}$

(D)

$\frac{k_z}{\mu_0 \omega} E_0^2 \sin^2(\omega t - k_z z) \hat{k}$

Detailed Solution

From Maxwell's equations, $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$ . Since $\vec{E} = E_0 \sin(\omega t - k_z z)\hat{j}$ , we have $\nabla \times \vec{E} = \frac{\partial E_y}{\partial z} \hat{i} = -(-k_z) E_0 \cos(\omega t - k_z z) \hat{i} = k_z E_0 \cos(\omega t - k_z z) \hat{i}$ . Integrating with respect to $t$ , $\vec{B} = -\frac{k_z E_0}{\omega} \sin(\omega t - k_z z) \hat{i}$ . The Poynting vector is $\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B}) = \frac{1}{\mu_0} [E_0 \sin(\omega t - k_z z) \hat{j} \times (- \frac{k_z E_0}{\omega} \sin(\omega t - k_z z) \hat{i})] = \frac{1}{\mu_0} \frac{k_z E_0^2}{\omega} \sin^2(\omega t - k_z z) \hat{k}$ .

Question 705035

Detailed Solution

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