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QPage 339: Q7: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Waves Page 339: Q7 Solution: The magnetic field in a plane electromagnetic wave is given by…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the peak energy density ($\frac{1}{2}\epsilon_0 E_0^2$) with the time-averaged energy density, which requires an additional factor of $1/2$ from the mean of $\sin^2$, resulting in $u_{avg} = \frac{1}{4}\epsilon_0 E_0^2$.

The magnetic field in a plane electromagnetic wave is given by B=200 μT)sin⁡[4.0×1015 s−1)t−x/c)]B = 200 \ \mu\text{T}) \sin [4.0 \times 10^{15} \text{ s}^{-1})t - x/c)]. Find the maximum electric field and the average energy density corresponding to the electric field.

Step-by-Step Verified Solution

**Answer:** E0=60 kV/mE_0 = 60 \, \text{kV/m}, average energy density =1.6×10−2 J/m3= 1.6 \times 10^{-2} \, \text{J/m}^3. **Step-by-Step Explanation:** Given: B=(200 μT)sin⁡[(4.0×1015 s−1)(t−x/c)]B = (200 \, \mu\text{T}) \sin[(4.0 \times 10^{15} \, \text{s}^{-1})(t - x/c)]. Maximum magnetic field: B0=200 μT=2×10−4 TB_0 = 200 \, \mu\text{T} = 2 \times 10^{-4} \, \text{T}. E0=cB0=3×108×2×10−4=6×104 V/m=60 kV/mE_0 = cB_0 = 3 \times 10^8 \times 2 \times 10^{-4} = 6 \times 10^4 \, \text{V/m} = 60 \, \text{kV/m}. Average energy density of electric field: uE=14ϵ0E02=14×8.85×10−12×(6×104)2=1.6×10−2 J/m3u_E = \frac{1}{4}\epsilon_0 E_0^2 = \frac{1}{4} \times 8.85 \times 10^{-12} \times (6 \times 10^4)^2 = 1.6 \times 10^{-2} \, \text{J/m}^3 [citation:4].

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