Question

The electric field in an electromagnetic wave is given by $\vec{E} = \hat{i} 40 \cos \omega \left( t - \frac{z}{c} \right) \, \text{NC}^{-1}.$The magnetic field induction of this wave is (in SI unit):

• A. $\vec{B} = \hat{i} \frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right)$
• B. $\vec{B} = \hat{j} 40 \cos \omega \left( t - \frac{z}{c} \right)$
• C. $\vec{B} = \hat{k} \frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right)$
• D. $\vec{B} = \hat{j} \frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right)$

Answer 1

Answer: (D)

$\vec{B} = \hat{j} \frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right)$

Answer 2

Given the electric field of the electromagnetic wave:
$\vec{E} = \hat{i} 40 \cos \omega \left( t - \frac{z}{c} \right) \, \text{NC}^{-1}$
In an electromagnetic wave, the magnetic field $\vec{B}$ is perpendicular to both the electric field $\vec{E}$ and the direction of propagation.

Since $\vec{E}$ is along the $\hat{i}$-direction and the wave propagates along the $\hat{k}$-direction, the magnetic field $\vec{B}$ must be along the $\hat{j}$-direction.

The relationship between the magnitudes of the electric and magnetic fields in an electromagnetic wave is given by:
$B = \frac{E}{c}$

Substituting the given electric field magnitude:
$B = \frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right)$
Thus, the magnetic field is:
$\vec{B} = \hat{j} \frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right)$