Question

A plane electromagnetic wave, has frequency of 2.0 × $10^{10}$ Hz and its energy density is 1.02 × $10^{-8}$ J/m³ in vacuum. The amplitude of the magnetic field of the wave is close to

• A. 190 nT
• B. 160 nT
• C. 150 nT
• D. 180 nT

Answer 1

Answer: (B)

160 nT

Answer 2

The average energy density ($u$) of a plane electromagnetic wave is given by $u = \frac{1}{2\mu_0} B_0^2$, where $B_0$ is the amplitude of the magnetic field. We are given $u = 1.02 \times 10^{-8}$ J/m³. Using the given constants $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$ Nm²/C² and $c = 3 \times 10^8$ m/s, we find $\mu_0 = 4\pi \times 10^{-7}$ H/m. Now, we solve for $B_0$: $B_0 = \sqrt{2\mu_0 u} = \sqrt{2 \times (4\pi \times 10^{-7} \text{ H/m}) \times (1.02 \times 10^{-8} \text{ J/m}^3)}$ $B_0 \approx \sqrt{25.635 \times 10^{-15}} \approx 1.601 \times 10^{-7}$ T. Converting to nanoTesla: $B_0 \approx 1.601 \times 10^{-7} \times 10^9$ nT $= 160.1$ nT. This is closest to 160 nT.