JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
Intensity of sunlight is observed as 0.092 Wm 2 at a point in free space. What will be the peak value of magnetic field at the point? ()
- A2.77 10 8 T
- B1.96 10 8 T
- C8.31 T
- D5.88 T
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Correct answer: A
- Use the relation between intensity and field amplitudes
For an electromagnetic wave in free space, the average intensity is
Also, since
we can write
=\frac{1}{2}\varepsilon_0 c^3 B_0^2$$ So, $$B_0=\sqrt{\frac{2I}{\varepsilon_0 c^3}}$$ --- 2. **Substitute the given values** Given: $$I=0.092\ \text{W m}^{-2}$$ $$\varepsilon_0=8.85\times 10^{-12}$$ $$c=3\times 10^8\ \text{m s}^{-1}$$ Thus, $$B_0=\sqrt{\frac{2\times 0.092}{(8.85\times 10^{-12})(3\times 10^8)^3}}$$ --- 3. **Compute the denominator** First, $$(3\times 10^8)^3=27\times 10^{24}=2.7\times 10^{25}$$ Now, $$\varepsilon_0 c^3=(8.85\times 10^{-12})(2.7\times 10^{25})$$ $$=23.895\times 10^{13}=2.3895\times 10^{14}$$ --- 4. **Compute the fraction** Numerator: $$2I=0.184$$ Therefore, $$B_0^2=\frac{0.184}{2.3895\times 10^{14}}\approx 7.70\times 10^{-16}$$ So, $$B_0=\sqrt{7.70\times 10^{-16}}\approx 2.77\times 10^{-8}\ \text{T}$$ --- 5. **Match with the options** $$B_0\approx 2.77\times 10^{-8}\ \text{T}$$ So the correct option is: **A: $2.77\times 10^{-8}\ \text{T}$** --- 6. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** They match.More from Electromagnetic Waves
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