Question

By measuring the heating of a material as it absorbs light from the sun, one finds that the intensity of sunlight at the surface of the earth is $1300 Wm ^{-2}$. What will be magnetic field of sunlight?

• A. $990 T$
• B. $3 \times 10^{-6} T$
• C. $3.3 \times 10^{-6} T$
• D. $9.9 \times 10^{-6} T$

Answer 1

Answer: (C)

Here, $I=1300 Wm ^{-2}$
As, $ I=\frac{1}{2} \varepsilon_{0} c E_{0}^{2}$
As, we know, electric field i.e., $E_{0}=\sqrt{\frac{2 I}{\varepsilon_{0} c}}$
$=\sqrt{\frac{2\left(1300 Wm ^{-2}\right)}{\left(8.85 \times 10^{-12} C ^{2} Nm ^{-2}\right)\left(3 \times 10^{8} ms ^{-1}\right)}}$
or $ E_{0}=990 NC ^{-1}$
So, magnetic field of sunlight $i . e$.,
$B_{0}=\frac{E_{0}}{c}=\frac{990 NC ^{-1}}{3 \times 10^{8} ms ^{-1}}=3.3 \times 10^{-6} T$