Laser Beam Electric Field Calculation Explained

This problem requires us to calculate the maximum electric field in an electromagnetic wave, specifically a laser beam, given its power, cross-sectional area, the permittivity of free space, and the speed of light. We will use the concepts of intensity of an electromagnetic wave, which relates power and area, and also relates intensity to the maximum electric field.

Given Parameters for the Laser Beam

Let's first list all the given values in their standard SI units to ensure consistency in our calculations.

• Laser Power (P): The power of the laser beam is given as 54 mW.
  • $P = 54 \text{ mW} = 54 \times 10^{-3} \text{ W}$
• Cross-sectional Area (A): The area through which the laser beam passes is 5.0 mm$^2$.
  • $A = 5.0 \text{ mm}^2 = 5.0 \times (10^{-3} \text{ m})^2 = 5.0 \times 10^{-6} \text{ m}^2$
• Permittivity of Space ($\epsilon_0$): This is a fundamental constant for electromagnetic calculations.
  • $\epsilon_0 = 9 \times 10^{-12} \text{ SI units}$ (approximately $9 \times 10^{-12} \text{ F/m}$)
• Speed of Light (c): The speed at which electromagnetic waves travel in a vacuum.
  • $c = 3 \times 10^8 \text{ m/s}$

We can summarize these values in a table for clarity:

Calculating Electromagnetic Wave Intensity

The intensity (I) of an electromagnetic wave is defined as the power (P) per unit cross-sectional area (A).

The formula for intensity is: $$I = \frac{P}{A}$$

Now, let's substitute the given values for power and area into this formula:

$$I = \frac{54 \times 10^{-3} \text{ W}}{5.0 \times 10^{-6} \text{ m}^2}$$ $$I = \frac{54}{5.0} \times 10^{(-3 - (-6))} \text{ W/m}^2$$ $$I = 10.8 \times 10^3 \text{ W/m}^2$$ $$I = 10800 \text{ W/m}^2$$

So, the intensity of the laser beam is $10800 \text{ W/m}^2$.

Determining Maximum Electric Field ($E_{max}$)

The intensity of an electromagnetic wave is also related to the maximum electric field ($E_{max}$) and maximum magnetic field ($B_{max}$) of the wave. For the electric field, the relationship is given by:

$$I = \frac{1}{2} c \epsilon_0 E_{max}^2$$

Our goal is to find $E_{max}$, so we need to rearrange this formula to solve for $E_{max}$:

$$E_{max}^2 = \frac{2I}{c \epsilon_0}$$ $$E_{max} = \sqrt{\frac{2I}{c \epsilon_0}}$$

Now, substitute the calculated intensity (I) and the given values for c and $\epsilon_0$:

$$E_{max} = \sqrt{\frac{2 \times 10800 \text{ W/m}^2}{(3 \times 10^8 \text{ m/s}) \times (9 \times 10^{-12} \text{ F/m})}}$$ $$E_{max} = \sqrt{\frac{21600}{27 \times 10^{(8 - 12)}}}$$ $$E_{max} = \sqrt{\frac{21600}{27 \times 10^{-4}}}$$ $$E_{max} = \sqrt{\frac{21600 \times 10^4}{27}}$$ $$E_{max} = \sqrt{800 \times 10^4}$$ $$E_{max} = \sqrt{8 \times 10^2 \times 10^4}$$ $$E_{max} = \sqrt{8 \times 10^6}$$

To simplify the square root, we can write $8 \times 10^6$ as $4 \times 2 \times 10^6$:

$$E_{max} = \sqrt{4 \times 2 \times 10^6}$$ $$E_{max} = 2 \times \sqrt{2} \times 10^3$$

Using the approximation $\sqrt{2} \approx 1.414$:

$$E_{max} = 2 \times 1.414 \times 10^3 \text{ V/m}$$ $$E_{max} = 2.828 \times 10^3 \text{ V/m}$$ $$E_{max} = 2828 \text{ V/m}$$

Since the options are given in kV/m, we convert V/m to kV/m by dividing by 1000:

$$E_{max} = \frac{2828}{1000} \text{ kV/m}$$ $$E_{max} = 2.828 \text{ kV/m}$$

Rounding to two decimal places, the magnitude of the maximum electric field is $2.82 \text{ kV/m}$.