Question

A light beam traveling in the x-direction is described by the magnetic field \( B_z = 2 \times 10^{-6} \sin \omega(t - x/c) \) Tesla. The value of the maximum electric field is

• A. 200 V/m
• B. 300 V/m
• C. 600 V/m
• D. 800 V/m

Hint:

For any electromagnetic wave in a vacuum, the ratio of the electric field amplitude to the magnetic field amplitude is always the speed of light: \( E_0 / B_0 = c \).

Answer 1

The Correct Option is C

Step 1: State the relationship between the amplitudes of electric and magnetic fields in an electromagnetic wave. In a vacuum (or air, approximately), the magnitudes of the electric field (\(E\)) and magnetic field (\(B\)) at any instant are related by: \[ E = cB \] where \(c\) is the speed of light. This relationship also holds for their maximum values (amplitudes), \(E_0\) and \(B_0\). \[ E_0 = c B_0 \]

Step 2: Identify the maximum magnetic field (\(B_0\)) from the given equation. The equation for the magnetic field is given in the form \( B_z = B_0 \sin(\dots) \). By comparing, we can see that the amplitude (maximum value) of the magnetic field is \( B_0 = 2 \times 10^{-6} \) T.

Step 3: Calculate the maximum electric field (\(E_0\)). Use the value of the speed of light, \( c \approx 3 \times 10^8 \) m/s. \[ E_0 = (3 \times 10^8 \text{ m/s}) \times (2 \times 10^{-6} \text{ T}) \] \[ E_0 = 6 \times 10^2 \text{ V/m} = 600 \text{ V/m} \]