The magnitude of the electric field of a plane electromagnetic wave travelling in free space is \( E \). If \( \mu_0 \) and \( \varepsilon_0 \) are respectively permeability and permittivity of the free space, then the magnitude of magnetic field of the wave is
Show Hint
- \( E \mu_0 \varepsilon_0 \)
- \( \frac{E}{\mu_0 \varepsilon_0} \)
- \( \frac{E}{\sqrt{\mu_0/\varepsilon_0}} \)
- \( \frac{E}{\sqrt{\mu_0 \varepsilon_0}} \)
The Correct Option is C
Solution and Explanation
Step 1: Relationship between electric and magnetic fields in an EM wave.
In a plane electromagnetic wave traveling through free space: \[ \frac{E}{B} = c \] where \( c \) is the speed of light.
Step 2: Express speed of light in terms of \( \mu_0 \) and \( \varepsilon_0 \). \[ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \Rightarrow B = \frac{E}{c} = E \cdot \sqrt{\mu_0 \varepsilon_0} \]
Step 3: Use identity involving magnetic field. We rewrite \( \sqrt{\mu_0 \varepsilon_0} = \sqrt{\mu_0 / (1/\varepsilon_0)} = \frac{1}{\sqrt{\mu_0/\varepsilon_0}} \)
Step 4: Final expression. \[ B = \frac{E}{\sqrt{\mu_0/\varepsilon_0}} \] which matches option (3).
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