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QPage 339: Q6: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Waves Page 339: Q6 Solution: The sunlight reaching Earth has maximum electric field of .…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently replace the correct relation $B_0=E_0/c$ with $B_0=E_0/Z_0$ (using the free‑space impedance $Z_0\approx377\,\Omega$), mistakenly treating $B$ (Tesla) as the magnetic field intensity $H$ (A/m). Remember that for a plane wave in vacuum $E=cB$, so divide by $c$, not by $Z_0$, to obtain the magnetic‑field amplitude.

The sunlight reaching Earth has maximum electric field of 810 V m−1810 \text{ V m}^{-1}. What is the maximum magnetic field in this light?

Step-by-Step Verified Solution

**Answer:** B0=2.7×10−6 TB_0 = 2.7 \times 10^{-6} \, \text{T}. **Step-by-Step Explanation:** Given: E0=810 V/mE_0 = 810 \, \text{V/m}, c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}. For an electromagnetic wave, B0=E0c=8103×108=2.7×10−6 TB_0 = \frac{E_0}{c} = \frac{810}{3 \times 10^8} = 2.7 \times 10^{-6} \, \text{T} [citation:4].

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