Question:
Intensity of sunlight is observed as \(0.092\ \mathrm{Wm^{-2}}\) at a point in free space. What will be the peak value of magnetic field at the point?
(\(\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2N^{-1}m^{-2}}\))
Intensity of sunlight is observed as \(0.092\ \mathrm{Wm^{-2}}\) at a point in free space. What will be the peak value of magnetic field at the point?
(\(\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2N^{-1}m^{-2}}\))
(\(\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2N^{-1}m^{-2}}\))
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Use \(I = \tfrac12 \varepsilon_0 c E_0^2\) and \(B_0 = E_0/c\), with \(c = 3\times10^8\ \mathrm{m/s}\).
Updated On: Nov 4, 2026
- \(2.77 \times 10^{-8}\ \mathrm{T}\)
- \(1.96 \times 10^{-8}\ \mathrm{T}\)
- \(8.31\ \mathrm{T}\)
- \(5.88\ \mathrm{T}\)
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The Correct Option is A
Solution and Explanation
Intensity \(I = \tfrac12 \varepsilon_0 c E_0^2\), so \(E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}}\).
\(E_0 = \sqrt{\dfrac{2 \times 0.092}{8.85\times10^{-12} \times 3\times10^{8}}} = \sqrt{\dfrac{0.184}{2.655\times10^{-3}}} = \sqrt{69.3} \approx 8.32\ \mathrm{V/m}\).
\(B_0 = \dfrac{E_0}{c} = \dfrac{8.32}{3\times10^{8}} \approx 2.77\times10^{-8}\ \mathrm{T}\).
Answer: (1) \(2.77 \times 10^{-8}\ \mathrm{T}\)
\(E_0 = \sqrt{\dfrac{2 \times 0.092}{8.85\times10^{-12} \times 3\times10^{8}}} = \sqrt{\dfrac{0.184}{2.655\times10^{-3}}} = \sqrt{69.3} \approx 8.32\ \mathrm{V/m}\).
\(B_0 = \dfrac{E_0}{c} = \dfrac{8.32}{3\times10^{8}} \approx 2.77\times10^{-8}\ \mathrm{T}\).
Answer: (1) \(2.77 \times 10^{-8}\ \mathrm{T}\)
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