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QPage 339: Q8: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Waves Page 339: Q8 Solution: A laser beam has intensity . Find amplitudes of electric and…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the time-averaged intensity formula $I = \frac{1}{2}\epsilon_0 c E_0^2$ with the instantaneous or RMS-based form, leading to a factor-of-two error in the amplitude calculation. Always verify whether the problem asks for peak amplitude ($E_0$) or RMS value ($E_{rms}$), and ensure the factor of $\frac{1}{2}$ is correctly included when deriving $E_0$ from $I$.

A laser beam has intensity 2.5×1014 W m−22.5 \times 10^{14} \text{ W m}^{-2}. Find amplitudes of electric and magnetic fields in the beam.

Step-by-Step Verified Solution

**Answer:** E0=4.3×108 V/mE_0 = 4.3 \times 10^8 \, \text{V/m}, B0=1.4 TB_0 = 1.4 \, \text{T}. **Step-by-Step Explanation:** Given: Intensity I=2.5×1014 W/m2I = 2.5 \times 10^{14} \, \text{W/m}^2. For an electromagnetic wave, I=12ϵ0cE02I = \frac{1}{2}\epsilon_0 c E_0^2, so E0=2Iϵ0cE_0 = \sqrt{\frac{2I}{\epsilon_0 c}}. E0=2×2.5×10148.85×10−12×3×108=5×10142.655×10−3=1.88×1017=4.3×108 V/mE_0 = \sqrt{\frac{2 \times 2.5 \times 10^{14}}{8.85 \times 10^{-12} \times 3 \times 10^8}} = \sqrt{\frac{5 \times 10^{14}}{2.655 \times 10^{-3}}} = \sqrt{1.88 \times 10^{17}} = 4.3 \times 10^8 \, \text{V/m}. B0=E0c=4.3×1083×108=1.4 TB_0 = \frac{E_0}{c} = \frac{4.3 \times 10^8}{3 \times 10^8} = 1.4 \, \text{T} [citation:4].

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