Question

If a laser beam has an intensity of \(2.5\times10^{14}\,\text{W/m}^{2}\), then the amplitude of the electric field and magnetic field in the beam is:

• A. \(4.3\times10^{8}\,\text{N/C},\;1.44\,\text{T}\)
• B. \(3\times10^{8}\,\text{N/C},\;1.44\,\text{T}\)
• C. \(43\times10^{8}\,\text{N/C},\;1.44\,\text{T}\)
• D. \(4.3\times10^{8}\,\text{N/C},\;14.4\,\text{T}\)

Hint:

Use I = (1/2) c eps0 E0^2 for E0, then B0 = E0/c.

Answer 1

The Correct Option is A

Step 1: The time-averaged intensity of an electromagnetic wave in terms of the electric field amplitude \(E_0\) is

\[I=\frac{1}{2}c\varepsilon_0 E_0^{2}.\]

Step 2: Solve for \(E_0\):

\[E_0=\sqrt{\frac{2I}{c\varepsilon_0}}=\sqrt{\frac{2(2.5\times10^{14})}{(3\times10^{8})(8.85\times10^{-12})}}.\]

Step 3: Evaluate. The denominator \(c\varepsilon_0=2.655\times10^{-3}\), so

\[E_0=\sqrt{\frac{5.0\times10^{14}}{2.655\times10^{-3}}}=\sqrt{1.88\times10^{17}}=4.3\times10^{8}\,\text{N/C}.\]

Step 4: The magnetic field amplitude is

\[B_0=\frac{E_0}{c}=\frac{4.3\times10^{8}}{3\times10^{8}}=1.44\,\text{T}.\]

This matches option (A).

\[\boxed{E_0\approx4.3\times10^{8}\,\text{N/C},\quad B_0\approx1.44\,\text{T}}\]