Question

In an electromagnetic wave in free space the root mean square value of the electric field is E_rms = 6 V/m. The peak value of the magnetic field is

• A. 1.41 × 10^–8 T
• B. 2.83 × 10^–8 T
• C. 0.70 × 10^–8T
• D. 4.23 × 10^–8T

Answer 1

The Correct Option is B

To find the peak value of the magnetic field in an electromagnetic wave in free space given the root mean square (RMS) value of the electric field, we can use the following relation:

E_0 = E_{\text{rms}} \times \sqrt{2}

where:

• E_0 is the peak value of the electric field.
• E_{\text{rms}} is given as 6 V/m.

Now, calculate the peak value of the electric field:

E_0 = 6 \, \text{V/m} \times \sqrt{2} \approx 6 \times 1.414 = 8.484 \, \text{V/m}

In free space, the speed of light (c) is related to the electric field (E_0) and the magnetic field (B_0) by the equation:

c = \frac{E_0}{B_0}

The speed of light (c) is approximately 3 \times 10^8 \, \text{m/s}. So, the peak value of the magnetic field (B_0) can be calculated as:

B_0 = \frac{E_0}{c} = \frac{8.484 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}}

Calculate B_0:

B_0 = \frac{8.484}{3 \times 10^8} \approx 2.83 \times 10^{-8} \, \text{T}

Therefore, the peak value of the magnetic field is 2.83 × 10^–8 T, matching option 2.