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QPage 339: Q5: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Waves Page 339: Q5 Solution: Using , find the ratio for a plane electromagnetic wave…

⚠️ Common Exam Traps & Conceptual Pitfalls

When converting between magnetic flux density $B$ and magnetic field strength $H$ in vacuum, students often drop the $mu_0$ factor. Since $E_0=cB_0$ and $B_0=mu_0H_0$, the correct ratio is $E_0/H_0=mu_0c$, not merely $c$. Omitting $mu_0$ gives an incorrect impedance value and wrong units.

Using B=μ0HB = \mu_0 H, find the ratio E0/H0E_0/H_0 for a plane electromagnetic wave propagating through vacuum. Show that it has the dimensions of electric resistance. This ratio is a universal constant called the impedance of free space.

Step-by-Step Verified Solution

**Answer:** E0H0=μ0c=μ0ϵ0≈377 Ω\frac{E_0}{H_0} = \mu_0 c = \sqrt{\frac{\mu_0}{\epsilon_0}} \approx 377 \, \Omega. This has dimensions of electric resistance. **Step-by-Step Explanation:** For an electromagnetic wave in vacuum, E0=cB0E_0 = cB_0 and B0=μ0H0B_0 = \mu_0 H_0. Thus, E0H0=cB0B0/μ0=μ0c\frac{E_0}{H_0} = \frac{cB_0}{B_0/\mu_0} = \mu_0 c. Since c=1μ0ϵ0c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}, μ0c=μ0ϵ0\mu_0 c = \sqrt{\frac{\mu_0}{\epsilon_0}}. Dimensions: [μ0]=[MLT−2A−2][\mu_0] = [MLT^{-2}A^{-2}], [c]=[LT−1][c] = [LT^{-1}], so [μ0c]=[ML2T−3A−2][\mu_0 c] = [ML^2T^{-3}A^{-2}], which is the dimension of resistance (ohm). This value is approximately 377 Ω\Omega [citation:4].

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