A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is \( E_y = 9.3 \, \text{V/m} \). Then, the magnetic field vector of the wave at that point is:
Show Hint
- \( B_z = 9.3 \times 10^{-8} \, \text{T} \)
- \( B_z = 1.55 \times 10^{-8} \, \text{T} \)
- \( B_z = 6.2 \times 10^{-8} \, \text{T} \)
- \( B_z = 3.1 \times 10^{-8} \, \text{T} \)
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to find the magnetic field vector \( B_z \) of a plane electromagnetic wave given its electric field vector \( E_y \) and the frequency of the wave.
First, recall that in electromagnetic waves traveling in free space, the electric field \( \mathbf{E} \) and magnetic field \( \mathbf{B} \) are related by the following relation:
\(E = cB\)
where:
- \(E = 9.3 \, \text{V/m}\) (Given electric field amplitude)
- \(c = 3 \times 10^8 \, \text{m/s}\) (Speed of light in free space)
- \(B\) is the magnetic field amplitude we need to find.
Step 1: Use the relation \(E = cB\), we solve for \(B\):
\(B = \frac{E}{c}\)
Step 2: Substitute the given values:
\(B = \frac{9.3}{3 \times 10^8}\)
Step 3: Calculate the magnetic field amplitude:
\(B = 3.1 \times 10^{-8} \, \text{T}\)
Therefore, the magnetic field vector of the wave at that point is:
Correct Answer: \(B_z = 3.1 \times 10^{-8} \, \text{T}\)
This matches option 4, confirming our calculation is correct. The problem uses the direct relationship between the electric and magnetic fields in electromagnetic waves, which is a fundamental principle of wave propagation in physics.
Approach Solution -2
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