JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2026
- Medium
- Numerical
The equation of the electric field of an electromagnetic wave propagating through free space is given by :
E \(=\sqrt{377} \sin \left(6.27 \times 10^3 t -2.09 \times 10^{-5} x \right)\) N/C
he average power of the electromagnetic wave is\(\left(\frac{1}{\alpha}\right) W / m ^2\). The value of \(\alpha\) is ___________
\(\left(\right.\) Take \(\sqrt{\frac{\mu_0}{\varepsilon_0}}=377\) in SI units \()\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
Here, \(v=\frac{\omega}{K}=\frac{6.27 \times 10^3}{2.09 \times 10^{-5}}\)
\(=3 \times 10^8\)
So, wave moving in vacuum
Now, \(I =\left(\frac{1}{2} \varepsilon_0 E _0^2\right) c =\frac{1}{2} \varepsilon_0 E _0^2 \frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
\(=\frac{1}{2} \sqrt{\frac{\varepsilon_0}{\mu_0}} E _0^2=\frac{1}{2} \frac{1}{377} \times 377\)
\(\frac{1}{d}=\frac{1}{2}\)
\(d =2\)
\(=3 \times 10^8\)
So, wave moving in vacuum
Now, \(I =\left(\frac{1}{2} \varepsilon_0 E _0^2\right) c =\frac{1}{2} \varepsilon_0 E _0^2 \frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
\(=\frac{1}{2} \sqrt{\frac{\varepsilon_0}{\mu_0}} E _0^2=\frac{1}{2} \frac{1}{377} \times 377\)
\(\frac{1}{d}=\frac{1}{2}\)
\(d =2\)
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