Question:

If the amplitude of the magnetic field in a travelling plane electromagnetic wave is \(2.2 \times 10^{-4} \, \text{T}\), then the intensity of the wave is nearly:

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For EM waves, intensity \(I = c B_0^2 / 2\mu_0\) relates magnetic field amplitude to energy flux.
Updated On: Jul 18, 2026
  • \(5.8 \times 10^6 \, \text{W/m}^2\)
  • \(4.2 \times 10^6 \, \text{W/m}^2\)
  • \(1.2 \times 10^7 \, \text{W/m}^2\)
  • \(8.8 \times 10^5 \, \text{W/m}^2\)
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The Correct Option is A

Solution and Explanation

Step 1: Recall intensity formula in terms of magnetic field.
\[ I = \frac{c B_0^2}{2 \mu_0} \]
where \(c = 3 \times 10^8 \, \text{m/s}, \, \mu_0 = 4 \pi \times 10^{-7} \, \text{H/m}, \, B_0 = 2.2 \times 10^{-4} \, \text{T}\).

Step 2: Substitute values.
\[ I = \frac{3 \times 10^8 \cdot (2.2 \times 10^{-4})^2}{2 \cdot 4 \pi \times 10^{-7}} \]

Step 3: Compute square of B.
\[ B_0^2 = (2.2 \times 10^{-4})^2 = 4.84 \times 10^{-8} \]

Step 4: Multiply by c and divide by \(2 \mu_0\).
\[ I \approx \frac{3 \times 10^8 \cdot 4.84 \times 10^{-8}}{2 \cdot 4 \pi \times 10^{-7}} \approx 5.8 \times 10^6 \, \text{W/m}^2 \]

Step 5: Verify units.
Units consistent as \(\text{W/m}^2\).

Step 6: Final conclusion.
\[ \boxed{5.8 \times 10^6 \, \text{W/m}^2} \]
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