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Step-by-Step Solution
Step 1: Identify the Given Quantities
• The frequency of the wave, $ f = 2.0 \times 10^{10} \text{ Hz} $.
• The energy density of the electromagnetic wave in vacuum, $ \frac{dU}{dV} = 1.02 \times 10^{-8} \, \text{J}/\text{m}^3 $.
• Speed of light, $ c = 3 \times 10^{8} \text{ m/s} $.
• Permittivity of free space, $ \varepsilon_0 $ (related through $ \frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2 $).
• We want to find the amplitude of the magnetic field, $ B_0 $.
Step 2: Write the Expression for Energy Density in Terms of the Magnetic Field
For an electromagnetic wave in vacuum, the energy density can be expressed as:
$ \frac{dU}{dV} = \frac{B_0^2}{2 \mu_0}. $
Here, $ B_0 $ is the amplitude of the magnetic field, and $ \mu_0 $ is the permeability of free space.
Step 3: Substitute the Known Value for the Energy Density
Given $ \frac{dU}{dV} = 1.02 \times 10^{-8} \, \text{J}/\text{m}^3 $, we have:
$ 1.02 \times 10^{-8} = \frac{B_0^2}{2 \mu_0}. $
Step 4: Express $ \mu_0 $ in Terms of $ c $ and $ \varepsilon_0 $
Recall the relation between $ \mu_0 $, $ \varepsilon_0 $, and the speed of light $ c $:
$ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \quad \Rightarrow \quad \mu_0 = \frac{1}{c^2 \varepsilon_0}. $
Step 5: Solve for $ B_0^2 $
Substituting $ \mu_0 = \frac{1}{c^2 \varepsilon_0} $ into the equation from Step 3:
$ 1.02 \times 10^{-8} = \frac{B_0^2}{2 \cdot \frac{1}{c^2 \varepsilon_0}}. $
Rearranging to solve for $ B_0^2 $:
$ B_0^2 = 1.02 \times 10^{-8} \times 2 \times \frac{1}{c^2 \varepsilon_0}. $
Step 6: Numerical Substitution
Using $ \frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 $ and $ c = 3 \times 10^8 \, \text{m/s} $, one finds
$ \varepsilon_0 = \frac{1}{4\pi} \times \frac{1}{9 \times 10^9} $ and hence $ \frac{1}{c^2 \varepsilon_0} \approx \frac{4\pi \times 9 \times 10^9}{9 \times 10^{16}} $.
Substituting these into the expression for $ B_0^2 $ and simplifying yields:
$ B_0^2 \approx 1.02 \times 10^{-8} \times 2 \times \left(\frac{4\pi \times 9 \times 10^9}{9 \times 10^{16}}\right). $
Step 7: Calculate $ B_0 $
After numerical simplification, we get:
$ B_0 \approx 1.6 \times 10^{-7} \text{ T}, $
which is
$ B_0 \approx 160 \text{ nT}. $
Final Answer
The amplitude of the magnetic field of the wave is approximately 160 nT.
