JEE Main 12 Jan 2019 Evening Question 18
Question: The mean intensity of radiation on the surface of the Sun is about $ 10^{8}W/m^{2} $ . The rms value of the corresponding magnetic field is closest to
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ {10^{-2}}T $
B) $ 1T $
C) $ {10^{-4}}T $
D) $ 10^{2}T $
Show Answer
Answer:
Correct Answer: C
Solution:
- Mean intensity $ =\frac{B_rms^{2}}{{\mu_0}}c $
$ \Rightarrow $ $ B_rms^{2}=\frac{10^{8}\times 4\pi \times {10^{-7}}}{3\times 10^{8}} $ $ B_rms^{{}}\approx {10^{-4}}T $
