Question

If electric field component is $E = 377\sin(\omega t + kx)$ V/m of an electromagnetic wave and \[ \sqrt{\frac{\mu_0}{\varepsilon_0}} = 377, \] then find the average intensity of the wave (in W/m$^2$):

• A. $188.5$
• B. $200$
• C. $100$
• D. $300$

Hint:

For electromagnetic waves in free space, remember the shortcut: $I_{\text{avg}}=\dfrac{E_0^2}{2Z}$ with $Z=377\,\Omega$.

Answer 1

The Correct Option is A

Concept: The average intensity of an electromagnetic wave is given by: \[ I_{\text{avg}} = \frac{E_0^2}{2Z} \] where $E_0$ is the amplitude of the electric field and $Z = \sqrt{\dfrac{\mu_0}{\varepsilon_0}}$ is the impedance of free space.
Step 1: Identify given quantities From the equation: \[ E = 377\sin(\omega t + kx) \] \[ E_0 = 377\ \text{V/m} \] Given: \[ Z = \sqrt{\frac{\mu_0}{\varepsilon_0}} = 377\ \Omega \]
Step 2: Use intensity formula \[ I_{\text{avg}} = \frac{E_0^2}{2Z} \]
Step 3: Substitute values \[ I_{\text{avg}} = \frac{(377)^2}{2\times 377} = \frac{377}{2} = 188.5\ \text{W/m}^2 \] Step 4: Hence, the average intensity of the wave is: \[ \boxed{188.5\ \text{W/m}^2} \]