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Step-by-Step Solution
Step 1: Identify the Direction of Wave Propagation
The problem states that the electromagnetic wave propagates along the positive Y-direction. In electromagnetic waves, the electric field vector (denoted by
$ \overrightarrow{E} $) and magnetic field vector (denoted by
$ \overrightarrow{B} $) are perpendicular to each other, and their cross product gives the direction of propagation of the wave.
Step 2: Relate the Directions of $ \overrightarrow{E} $, $ \overrightarrow{B} $, and Propagation
The direction of propagation is given by
$ \overrightarrow{E} \times \overrightarrow{B} $. Because the wave travels in the +Y direction, if we choose
$ \overrightarrow{E} $ along the +Z axis, then
$ \overrightarrow{B} $ must lie along the +X axis to satisfy
$ (\overrightarrow{E} \times \overrightarrow{B}) = \overrightarrow{k}_y \,.$
Specifically,
$ (\widehat{k} \times \widehat{i}) = \widehat{j} $. Here
$ \widehat{i}, \widehat{j}, \widehat{k} $ are the unit vectors along the X, Y, and Z axes respectively.
Step 3: Write the Wave Equations
An electromagnetic wave traveling in the +Y direction can be written using the argument
$ (y - ct) $. Hence, the electric field can be expressed as:
$ \overrightarrow{E}(y, t) = E_0 \cos\biggl(\frac{2 \pi}{\lambda}(y - ct)\biggr)\,\widehat{k} \,,$
and the corresponding magnetic field is:
$ \overrightarrow{B}(y, t) = B_0 \cos\biggl(\frac{2 \pi}{\lambda}(y - ct)\biggr)\,\widehat{i} \,.$
Below is the illustrative figure (maintained from the original solution):
Step 4: Relate the Electric and Magnetic Field Magnitudes
In free space, for an electromagnetic wave:
$ E_0 = c \, B_0 \,.$
Step 5: Use the Relation Between Intensity and Field Amplitudes
The intensity
$ I $ of an electromagnetic wave is given by:
$ I = \frac{1}{2} \, c \, \epsilon_0 \, E_0^2 \,, $
where
$ \epsilon_0 $ is the permittivity of free space and
$ c $ is the speed of light in vacuum. Solving for
$ E_0 $ gives:
$ E_0 = \sqrt{\frac{2 I}{c \,\epsilon_0}} \,.$
Since
$ E_0 = c\,B_0 ,$ we get:
$ B_0 = \frac{E_0}{c} = \frac{1}{c} \sqrt{\frac{2 I}{c \,\epsilon_0}} \,.$
Step 6: Final Expressions for $ \overrightarrow{E} $ and $ \overrightarrow{B} $
Putting these all together, the fields for the wave traveling in the +Y direction are:
$ \overrightarrow{E}(y,t) \;=\; \sqrt{\frac{2 I}{c \,\epsilon_0}} \,\cos\!\Bigl[\frac{2\pi}{\lambda}\,(y - ct)\Bigr]\;\widehat{k} \,, $
$ \overrightarrow{B}(y,t) \;=\; \frac{1}{c}\;\sqrt{\frac{2 I}{c \,\epsilon_0}}\;\cos\!\Bigl[\frac{2\pi}{\lambda}\,(y - ct)\Bigr]\;\widehat{i} \,. $
These expressions satisfy the required direction of propagation, the perpendicularity of
$ \overrightarrow{E} $ and
$ \overrightarrow{B} $, and the correct magnitude relation in free space.
