Question:
A plane electromagnetic wave, has frequency of \(2.0 \times 10^{10}\) Hz and its energy density is \(1.02 \times 10^{-8}\) J/m\(^3\) in vacuum. The amplitude of the magnetic field of the wave is close to
(\(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \frac{\mathrm{N\,m^2}}{\mathrm{C^2}}\) and speed of light
\(= 3 \times 10^8\ \mathrm{ms^{-1}}\))
A plane electromagnetic wave, has frequency of \(2.0 \times 10^{10}\) Hz and its energy density is \(1.02 \times 10^{-8}\) J/m\(^3\) in vacuum. The amplitude of the magnetic field of the wave is close to
(\(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \frac{\mathrm{N\,m^2}}{\mathrm{C^2}}\) and speed of light
\(= 3 \times 10^8\ \mathrm{ms^{-1}}\))
(\(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \frac{\mathrm{N\,m^2}}{\mathrm{C^2}}\) and speed of light
\(= 3 \times 10^8\ \mathrm{ms^{-1}}\))
Show Hint
Energy density of an EM wave is \(u = \frac{B_0^2}{2\mu_0}\) (electric and magnetic parts are equal); find \(\mu_0\) from \(\frac{1}{\mu_0\varepsilon_0} = c^2\).
Updated On: Nov 5, 2026
- 190 nT
- 150 nT
- 160 nT
- 180 nT
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The Correct Option is C
Solution and Explanation
\(\varepsilon_0 = \frac{1}{4\pi \times 9 \times 10^9}\), so \(\mu_0 = \frac{1}{\varepsilon_0 c^2} = \frac{4\pi \times 9 \times 10^9}{9 \times 10^{16}} = 4\pi \times 10^{-7}\) T m/A.
Energy density \(u = \frac{B_0^2}{2\mu_0}\), hence \(B_0 = \sqrt{2\mu_0 u}\).
\(B_0 = \sqrt{2 \times 4\pi \times 10^{-7} \times 1.02 \times 10^{-8}} = \sqrt{2.56 \times 10^{-14}} \approx 1.6 \times 10^{-7}\) T \(= 160\) nT.
Answer: (3) 160 nT
Energy density \(u = \frac{B_0^2}{2\mu_0}\), hence \(B_0 = \sqrt{2\mu_0 u}\).
\(B_0 = \sqrt{2 \times 4\pi \times 10^{-7} \times 1.02 \times 10^{-8}} = \sqrt{2.56 \times 10^{-14}} \approx 1.6 \times 10^{-7}\) T \(= 160\) nT.
Answer: (3) 160 nT
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