JEE PYQ: Electromagnetic Waves Question 11
Question 11 - 2020 (02 Sep 2020 Shift 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 ($\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}$ $\frac{Nm^2}{C^2}$ and speed of light $= 3 \times 10^{8}$ ms$^{-1}$):
(1) 150 nT
(2) 160 nT
(3) 180 nT
(4) 190 nT
Show Answer
Answer: (2)
Solution
Energy density $= \frac{1}{2} \frac{B^2}{\mu_0}$. $B = \sqrt{2 \times \mu_0 \times 1.02 \times 10^{-8}} = 160$ nT.
BETA
