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In an electromagnetic wave in free space the root mean square value of the electric field is Erms=6E_{rms} = 6 V/m. The peak value of the magnetic field is:

A

1.41×10−81.41 \times 10^{-8} T

B

2.83×10−82.83 \times 10^{-8} T

C

0.70×10−80.70 \times 10^{-8} T

D

4.23×10−84.23 \times 10^{-8} T

Step-by-Step Solution

  1. Relationship between fields: In an electromagnetic wave, the amplitudes of the electric field (E0E_0) and magnetic field (B0B_0) are related by the speed of light (cc): c=E0B0c = \frac{E_0}{B_0}. Similarly, the rms values are related by c=ErmsBrmsc = \frac{E_{rms}}{B_{rms}} .
  2. Calculate Peak Electric Field (E0E_0): The peak value is related to the rms value by E0=2ErmsE_0 = \sqrt{2} E_{rms}. E0=2×6 V/m≈1.414×6=8.484 V/mE_0 = \sqrt{2} \times 6 \text{ V/m} \approx 1.414 \times 6 = 8.484 \text{ V/m}
  3. Calculate Peak Magnetic Field (B0B_0): Using B0=E0cB_0 = \frac{E_0}{c} where c=3×108c = 3 \times 10^8 m/s . B0=8.4843×108=2.828×10−8 TB_0 = \frac{8.484}{3 \times 10^8} = 2.828 \times 10^{-8} \text{ T} Rounding to two decimal places, B0≈2.83×10−8B_0 \approx 2.83 \times 10^{-8} T. Alternatively, find Brms=Erms/c=2×10−8B_{rms} = E_{rms}/c = 2 \times 10^{-8} T, then B0=2Brms=1.414×2×10−8≈2.83×10−8B_0 = \sqrt{2} B_{rms} = 1.414 \times 2 \times 10^{-8} \approx 2.83 \times 10^{-8} T.
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