JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2020
- Hard
- Numerical
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\(....nT\) \(\left(\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{\circ} \frac{ Nm ^{2}}{ C ^{2}}\right.\) and speed of \(1 ight\) \(\left.=3 \times 10^{8}\, ms ^{-1}\right)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(160\)
Step-by-step Solution
Detailed explanation
Energy density \(\frac{ dU }{ dV }=\frac{ B _{0}^{2}}{2 \mu_{0}}\)
\(1.02 \times 10^{-8}=\frac{ B _{0}^{2}}{2 \times 4 \pi \times 10^{-7}}\)
\(B _{0}^{2}=\left(1.02 \times 10^{-8}\right) \times\left(8 \pi \times 10^{-7}\right)\)
\(B _{0}=16 \times 10^{-8} T =160 nT\)
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