 Solution

ID: past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/a9/solution

Approximate the beam as uniform over its circular cross section. Its mean intensity is

I = P / [π (D /2)^{2}] = 1.27 \times 1 0^{6} W m^{- 2}

I = c B_{rms}^{2} / μ_{0}

, so

B_{rms} = \frac{μ _{0} P}{c π ( D /2 ) ^{2}} = 7.3 \times 1 0^{- 5} T .

This is an rms field, not the peak sinusoidal amplitude. A Gaussian-beam peak would need a stated diameter convention.

 New to topics? Read the docs here!