Question

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is W/m². What is the rms value of the electric field in the electromagnetic wave emitted by the laser?

Answer 1

I think your question should be:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is

$S = 1.23*10^9 W/m^2$

What is the rms value of (a) the electric field and

(b) the magnetic field in the electromagnetic wave emitted by the laser

Answer:

a) $6.81*10^5 N/c$

b) $2.27*10^3 T$

Explanation:

To find the RMS value of the electric field, let's use the formula:

$E_r_m_s = sqrt*(S / CE_o)$

Where

$C = 3.00 * 10^-^8 m/s$;

$E_o = 8.85*10^-^1^2 C^2/N.m^2$;

$S = 1.23*10^9 W/m^2$

Therefore

$E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]}$

$E_r_m_s= 6.81 *10^5N/c$

b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:

$B_r_m_s = E_r_m_s / C$;

$= 6.81*10^5 N/c / 3*10^8m/s$;

$B_r_m_s = 2.27*10^3 T$

Answer 2

Complete Question:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is 1.38 * 10⁹ W/m². What is the rms value of the electric field in the electromagnetic wave emitted by the laser?

Answer:

E = 7.21 * 10⁶ N/C

Explanation:

S = 1.38 * 10⁹ W/m²..............(1)

The formula for the average intensity of light can be given by:

S = c∈₀E²

The speed of light, c = 3 * 10⁸ m/s²

Permittivity of air, ∈₀ = 8.85 * 10⁻¹²m-3 kg⁻¹s⁴ A²

Substituting these parameters into equation (1)

1.38 * 10⁹ = 3 * 10⁸ * 8.85 * 10⁻¹² * E²

E² = (1.38 * 10⁹)/(3 * 10⁸ * 8.85 * 10⁻¹²)

E² = 0.052 * 10¹⁷

E = 7.21 * 10⁶ N/C