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Step-by-Step Solution
Step 1: Identify the components of the given electric field
The electric field of the uniform plane wave is given by
$E = -301.6 \sin(kz - \omega t)\,\hat{a}_x + 452.4 \sin(kz - \omega t)\,\hat{a}_y.$
Hence, the amplitude components of the electric field are:
$E_{0x} = 301.6$ V/m (the magnitude of the x-component)
$E_{0y} = 452.4$ V/m (the magnitude of the y-component)
Step 2: Calculate the resultant amplitude of the electric field
The magnitude $E_0$ of the resultant electric field amplitude is found by combining the two perpendicular components:
$E_0 = \sqrt{\bigl(E_{0x}\bigr)^2 + \bigl(E_{0y}\bigr)^2}
= \sqrt{(301.6)^2 + (452.4)^2}.$
Numerically, this is approximately $E_0 \approx 543.7 \,\text{V/m}.$
Step 3: Relate the electric and magnetic fields in an electromagnetic wave
For an electromagnetic wave traveling in free space, the magnitudes of the electric field $E_0$ and the magnetic field $B_0$ are related by the speed of light $c$:
$B_0 = \frac{E_0}{c},
\quad \text{where } c = 3 \times 10^8 \,\text{m/s}.$
Substituting $E_0 \approx 543.7 \,\text{V/m}$ gives
$B_0 = \frac{543.7}{3 \times 10^8} \,\text{T} \approx 1.81 \times 10^{-6} \,\text{T}.$
Step 4: Determine the direction of the magnetic field
The electromagnetic wave is propagating along the $+z$ direction (as implied by the argument $(kz - \omega t)$). In a right-handed coordinate system, if
$\overrightarrow{k} \parallel \hat{z},$
and the electric field has components in the $x$–$y$ plane, then the magnetic field
$\overrightarrow{B}$ is obtained via
$\overrightarrow{B}_0 = \frac{1}{c}\,\bigl(\hat{z} \times \overrightarrow{E}_0\bigr).$
Since
$\overrightarrow{E}_0 = -301.6\,\hat{x} + 452.4\,\hat{y},$
we compute
$\hat{z} \times \bigl(-301.6\,\hat{x} + 452.4\,\hat{y}\bigr)
= -301.6\,(\hat{z} \times \hat{x}) + 452.4\,(\hat{z} \times \hat{y}).$
Using the right-hand rule:
$\hat{z} \times \hat{x} = +\hat{y}$
and
$\hat{z} \times \hat{y} = -\hat{x},$
we get
$\overrightarrow{B}_0 = -301.6\,\hat{y} \;-\; 452.4\,\hat{x}.$
Hence,
$\overrightarrow{B} = \overrightarrow{B}_0 \,\sin(kz - \omega t)
= -301.6 \,\sin(kz - \omega t)\,\hat{y}
\;-\; 452.4 \,\sin(kz - \omega t)\,\hat{x},$
scaled by $1/c$ overall. Numerically, the amplitudes must be divided by $c$ to get Tesla (T).
Step 5: Convert the magnetic flux density B to magnetic field intensity H
In free space,
$\overrightarrow{B} = \mu_0\,\overrightarrow{H},$
where
$\mu_0 = 4 \pi \times 10^{-7}\,\text{N\,A}^{-2}.$
Therefore,
$\overrightarrow{H} = \frac{\overrightarrow{B}}{\mu_0}
= \frac{1}{\mu_0}\Bigl(\frac{\overrightarrow{E}_0}{c}\Bigr)\sin(kz - \omega t)\times(\text{direction factor}).$
This leads to numerical values for each component:
$H_{0x} = \frac{-E_{0y}}{\mu_0\,c},
\quad
H_{0y} = \frac{-E_{0x}}{\mu_0\,c}.$
Substituting $E_{0x} = 301.6,$ $E_{0y} = 452.4,$ $c = 3 \times 10^8,$ and $\mu_0 = 4 \pi \times 10^{-7}$ yields approximately:
$H_{0x} \approx -1.2,\quad
H_{0y} \approx -0.8 \;\;(\text{in Am}^{-1}).$
Step 6: Write the final expression for the magnetic field intensity H
Hence, the magnetic field intensity is:
$\overrightarrow{H} = -0.8 \,\sin(kz - \omega t)\,\hat{a}_y \;-\; 1.2 \,\sin(kz - \omega t)\,\hat{a}_x\,\text{(A/m)}.$
This matches the given correct answer.
