Question

If the average power per unit area delivered by an electromagnetic wave is
\[ 9240\ \text{W m}^{-2} \] then the amplitude of the oscillating magnetic field in the EM wave is

• A. \(4.4\ \mu\text{T}\)
• B. \(6.6\ \mu\text{T}\)
• C. \(8.8\ \mu\text{T}\)
• D. \(10.2\ \mu\text{T}\)

Hint:

For electromagnetic waves, intensity is related to magnetic field amplitude by \(I=\dfrac{cB_0^2}{2\mu_0}\).

Answer 1

The Correct Option is C

Step 1: Relate intensity to the field amplitude.
The average power per unit area (intensity) of an EM wave in terms of the magnetic amplitude $B_0$ is \[ I = \frac{c\,B_0^{2}}{2\mu_0}. \]
Step 2: List the constants.
$I = 9240\ \text{W m}^{-2}$, $c = 3\times10^{8}\ \text{m s}^{-1}$, $\mu_0 = 4\pi\times10^{-7}\ \text{H m}^{-1}$.
Step 3: Solve for $B_0$.
\[ B_0 = \sqrt{\frac{2\mu_0 I}{c}}. \]
Step 4: Substitute the numbers.
\[ B_0 = \sqrt{\frac{2(4\pi\times10^{-7})(9240)}{3\times10^{8}}}. \]
Step 5: Simplify inside the root.
The numerator is $8\pi\times9240\times10^{-7} \approx 2.32\times10^{-2}$, and dividing by $3\times10^{8}$ gives \[ B_0 \approx \sqrt{7.74\times10^{-11}} \approx 8.8\times10^{-6}\ \text{T}. \]
Step 6: Conclude.
So the magnetic field amplitude is $8.8\ \mu\text{T}$.
\[ \boxed{8.8\ \mu\text{T}} \]