Question

Assume the helium-neon lasers commonly used in student physics laboratories have power outputs of 0.250mW. (a) If such a laser beam is projected onto a circular spot 1.00mm in diameter, what is its intensity? (b) Find the peak magnetic field strength. (c) Find the peak electric field strength.

Short answer

1. Intensity is$318.47\mathrm{W}/{\mathrm{m}}^2$.
2. The magnetic field strength is$16.32\times {10}^{-7}\mathrm{T}$.
3. The electric field strength is $4.89\times {10}^2\mathrm{V}/\mathrm{m}$.

Step-by-step solution

Define magnetic field strength

Magnetic field strength is a physical number that is one of the most fundamental measurements of the magnetic field's intensity.

Evaluating the intensity

(a)

Intensity is defined as power per unit area and may be calculated using the following formula:

$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}$

Where I is intensity,P is power and A is area.

Substitute the values in above equation,

$$\begin{gathered} \mathrm{I}=\frac{0.25\times {10}^{-3}\mathrm{W}}{\mathrm{\tau \tau }{\left(0.5\times {10}^{-3}\mathrm{m}\right)}^2} \\ =\frac{0.25\times {10}^{-3}\mathrm{W}}{0.785\times {10}^{-6}{\mathrm{m}}^2} \\ =318.47\mathrm{W}/{\mathrm{m}}^2 \\ \mathrm{Therefore},\mathrm{intensity}\mathrm{is}318.47\mathrm{W}/{\mathrm{m}}^2. \end{gathered}$$

Evaluating the magnetic field strength

(b)

The formula for intensity in terms of magnetic field and light speed is

$$\begin{gathered} \mathrm{I}=\frac{{\mathrm{cB}}_0^2}{2{\mathrm{\mu }}_0} \\ {\mathrm{B}}_0=\sqrt{\frac{2{\mathrm{\mu }}_0\mathrm{I}}{\mathrm{C}}} \\ \mathrm{Where}\mathrm{c}\mathrm{is}\mathrm{light}\mathrm{speed},{\mathrm{\mu }}_0\mathrm{is}\mathrm{permeability}\mathrm{of}\mathrm{free}\mathrm{space}\mathrm{and}\mathrm{I}\mathrm{is}\mathrm{intensity}. \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}, \\ {\mathrm{B}}_0=\sqrt{\frac{2\times \mathrm{\tau \tau }\times {10}^{-7}\mathrm{N}/\mathrm{A}\times 318.47\mathrm{W}/{\mathrm{m}}^2}{3\times {10}^8\mathrm{m}/\mathrm{s}}} \\ =16.32\times {10}^{-7}\mathrm{T} \\ \mathrm{Therefore},\mathrm{the}\mathrm{magnetic}\mathrm{field}\mathrm{strength}\mathrm{is}16.32\times {10}^{-7}\mathrm{T}. \end{gathered}$$

Evaluating the electric field strength

(c)

The equation describes the relationship between peak magnetic and peak electric fields is

$$\begin{gathered} B_0=\frac{E_0}{c} \\ {\mathrm{E}}_0=B_{0}c \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}, \\ {\mathrm{E}}_0=\left(16.32\times {10}^{-7}\mathrm{T}\right)\times \left(3\times {10}^8\mathrm{m}/\mathrm{s}\right) \\ =4.89\times {10}^2\mathrm{V}/\mathrm{m} \\ \mathrm{Therefore},\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{strength}\mathrm{is}4.89\times {10}^2\mathrm{V}/\mathrm{m}. \end{gathered}$$