Question

In an electromagnetic wave traveling in a vacuum, the electric field amplitude is \( 3.0 \times 10^3 \, \text{V/m} \). What is the magnetic field amplitude of the wave? Assume the speed of light in vacuum is \( c = 3.0 \times 10^8 \, \text{m/s} \).

• A. \( 1.0 \times 10^{-5} \, \text{T} \)
• B. \( 1.0 \times 10^{-3} \, \text{T} \)
• C. \( 1.0 \times 10^{-6} \, \text{T} \)
• D. \( 1.0 \times 10^{-4} \, \text{T} \)

Hint:

In an electromagnetic wave, the electric and magnetic field amplitudes are related by the speed of light, \( E_0 = c B_0 \). This relationship can be used to find one if the other is known.

Answer 1

The Correct Option is A

In an electromagnetic wave, the relationship between the electric field amplitude (\( E_0 \)) and the magnetic field amplitude (\( B_0 \)) is given by the equation: \[ E_0 = c B_0 \] Where: - \( E_0 = 3.0 \times 10^3 \, \text{V/m} \) (electric field amplitude), - \( c = 3.0 \times 10^8 \, \text{m/s} \) (speed of light in vacuum). Rearranging the equation to solve for \( B_0 \): \[ B_0 = \frac{E_0}{c} \] Substitute the known values: \[ B_0 = \frac{3.0 \times 10^3}{3.0 \times 10^8} = 1.0 \times 10^{-5} \, \text{T} \] Thus, the magnetic field amplitude is \( 1.0 \times 10^{-5} \, \text{T} \).