Question 45: EM waves

Question

The magnetic field in a plane electromagnetic wave is $$\mathrm{B}_{\mathrm{y}}=\left(3.5 	imes 10^{-7}ight) \sin \left(1.5 	imes 10^3 x+0.5 	imes 10^{11} tight) \mathrm{T}$$. The corresponding electric field will be :

Accepted Answer

For an electromagnetic wave propagating in free space, the relationship between the magnitudes of the electric field ($$E$$) and the magnetic field ($$B$$) can be described using the equation:
$E = cB$
where
$c$ is the speed of light in vacuum, approximately $3.0 	imes 10^8 \, 	ext{m/s}$.
$B$ is the magnitude of the magnetic field.
Given the magnetic field $B_y = (3.5 	imes 10^{-7}) \sin (1.5 	imes 10^3 x + 0.5 	imes 10^{11} t) \, 	ext{T}$, we can calculate the corresponding electric field magnitude using the formula above:
$E = (3.0 	imes 10^8) 	imes (3.5 	imes 10^{-7})$
$= 105 \, 	ext{Vm}^{-1}$
Thus, the magnitude of the electric field associated with the given magnetic field is $105 \, 	ext{Vm}^{-1}$. The direction of the electric field is perpendicular to both the magnetic field and the direction of propagation. Given $B_y$, this means $E$ will have components in the $x-z$ plane. Since electromagnetic waves are transverse, and given that the magnetic field is specified to be in the $y$-direction, the corresponding electric field component must lie in a plane perpendicular to the $y$-axis, which could be either the $x$ or the $z$ direction.
However, knowing electromagnetic wave properties, if the wave is propagating along the $x$-axis and the magnetic field ($B_y$) is along the $y$-axis, then by right-hand rule, the electric field ($E$) must be along the $z$-axis to maintain the orthogonal relationship among the direction of propagation, electric field, and magnetic field vector directions.
Therefore, the correct option is:
$	ext{Option A: } E_z = 105 \sin \left(1.5 	imes 10^3 x + 0.5 	imes 10^{11} tight) \, 	ext{Vm}^{-1}$

Answer

$$E_z=105 \sin \left(1.5 	imes 10^3 x+0.5 	imes 10^{11} tight) \mathrm{Vm}^{-1}$$

Answer

$$E_y=10.5 \sin \left(1.5 	imes 10^3 x+0.5 	imes 10^{11} tight) \mathrm{Vm}^{-1}$$

Answer

$$E_y=1.17 \sin \left(1.5 	imes 10^3 x+0.5 	imes 10^{11} tight) \mathrm{Vm}^{-1}$$

Answer

$$E_z=1.17 \sin \left(1.5 	imes 10^3 x+0.5 	imes 10^{11} tight) \mathrm{Vm}^{-1}$$

EM waves

The magnetic field in a plane electromagnetic wave is $\mathrm{B}_{\mathrm{y}}=\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{T}$. The corresponding electric field will be :

Choose the correct answer: