ANSWERS
(a) I = 50.92 W/m²
(b) B₀ = 6.53 x 10⁻⁷ T
(c) E₀ = 195.9 V/m
Step-by-step explanation
Given:
• the power output of a helium-neon laser, P = 0.250 mW = 2.5 x 10⁻⁴ W
Find:
• (a), The intensity, I in W/m², if a laser beam is projected into a circular spot with a diameter d = 2.50 mm
,
• (b), The peak magnetic field strength, B₀ in T
,
• (c), The peak electric field strength, E₀ in V/m
(a) The intensity of a beam of light of power P projected in area A is,
![I=(P)/(A)](https://img.qammunity.org/2023/formulas/physics/college/3ta1t4lrqt16vqwfd0wog5me4gghk6plbl.png)
We know that the circular spot has a diameter of 2.50 mm, so its radius is,
![r=(d)/(2)=(2.50mm)/(2)=1.25mm=1.25\cdot10^(-3)m](https://img.qammunity.org/2023/formulas/physics/college/pym5y70zi2702lqg7rteh10sqdil8ln32v.png)
So, the area of the circular spot is,
![A=\pi r^2=\pi\cdot(1.25\cdot10^(-3))^2m^2\approx4.91\cdot10^(-6)m^2](https://img.qammunity.org/2023/formulas/physics/college/6x46j0ifm98enx45r36u3nh77ok2yzu3ss.png)
And the intensity is,
![I=(P)/(A)=(2.5\cdot10^(-4)W)/(4.91\cdot10^(-6)m^2)\approx50.92W/m^2](https://img.qammunity.org/2023/formulas/physics/college/cok1gtews1me89pbbzs6xdg0avsm0qd0ce.png)
Hence, the intensity of the laser beam is 50.92 W/m², rounded to two decimal places.
(b) The intensity of the beam is also related to the peak magnetic field as follows,
![I=(c\cdot B_0^2)/(2\mu_0)](https://img.qammunity.org/2023/formulas/physics/college/jku2ldxmt7riej47zcbp04q7gr2x775ot8.png)
Where c is the speed of light, 3 x 10⁸ m/s, and μ₀ is the vacuum magnetic permeability, 1.257 x 10⁻⁶ H/m.
Solving this equation for B₀,
![B_0=\sqrt{(2I\mu_0)/(c)}=\sqrt{(2\cdot50.92W/m^2\cdot1.257\cdot10^(-6)H/m)/(3\cdot10^8m/s)}\approx6.53\cdot10^(-7)T](https://img.qammunity.org/2023/formulas/physics/college/8bdgmp41j76gusfc1del37wnjoxnzf2caa.png)
Hence, the peak magnetic field strength is 6.53 x 10⁻⁷ T.
(c) The peak electric field strength is proportional to the peak magnetic field strength. The constant of proportionality is the speed of light,
![E_0=cB_0](https://img.qammunity.org/2023/formulas/physics/college/oqqq29vhn3cu29iyl1a8u6ltvzj8zbq1mk.png)
Using the peak magnetic field strength found in part (b),
![E_0=3\cdot10^8m/s\cdot6.53\cdot10^(-7)T\approx195.9V/m](https://img.qammunity.org/2023/formulas/physics/college/wgzlpf3kmyy70fmb8toiq4mtsanjcv33nn.png)
Hence, the peak electric field strength is 195.9 V/m.