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Step-by-Step Solution
Step 1: Understand the Physical Setup
A plane electromagnetic wave of frequency 100 MHz is traveling along the x-direction. At a specific point in space and time, the magnetic field is given by
$ \overrightarrow{B} = 2.0 \times 10^{-8} \,\hat{k} \text{ T} $,
where $ \hat{k} $ is the unit vector along the z-direction. We need to find the corresponding electric field $ \overrightarrow{E} $ at that same point.
Step 2: Recall the Relationship between $ \overrightarrow{E} $ and $ \overrightarrow{B} $
For an electromagnetic wave traveling in vacuum:
The electric field $ \overrightarrow{E} $, magnetic field $ \overrightarrow{B} $, and direction of propagation (wave vector $ \overrightarrow{k} $) are mutually perpendicular.
The magnitudes of the fields are related by
$ E = c\,B $,
where $ c $ is the speed of light in vacuum (approximately $ 3 \times 10^{8} \,\text{m/s} $).
Step 3: Calculate the Magnitude of $ \overrightarrow{E} $
Given:
$ B = 2.0 \times 10^{-8} \,\text{T} $
Using the relationship,
$ E = c\,B: $
$ E = \bigl(3 \times 10^{8} \,\text{m/s}\bigr)\times \bigl(2.0 \times 10^{-8} \,\text{T}\bigr) = 6.0 \,\text{V/m}. $
Step 4: Determine the Direction of $ \overrightarrow{E} $
The wave travels along the x-direction ($ \hat{i} $), and $ \overrightarrow{B} $ is along the z-direction ($ \hat{k} $). Thus, by the right-hand rule for electromagnetic waves, $ \overrightarrow{E} $ must be perpendicular to both $ \hat{i} $ (the direction of propagation) and $ \hat{k} $ (the direction of $ \overrightarrow{B} $). Hence, $ \overrightarrow{E} $ points along the y-direction ($ \hat{j} $).
Step 5: Write the Final Electric Field Vector
The electric field vector is therefore
$ \overrightarrow{E} = 6.0 \,\hat{j} \,\text{V/m}. $
Final Answer
$ \overrightarrow{E} = 6.0 \,\hat{j} \,\text{V/m} $.
