Consider an electron revolving in a circular orbit of a hydrogen atom with an orbital speed v . The revolving electron produces a magnetic field B at the center of the nucleus. If the magnetic field B is proportional to v^x , what is the numerical value of the exponent x ?

Step-by-step solution

The electrostatic force provides the necessary centripetal force for the electron: mv^2 r = 1 4 ₀ e^2 r^2 Rearranging for r , we get: r = 1 4 ₀ e^2 mv^2 This shows that r 1 v^2 . The revolving electron constitutes an electric current I : I = e T = ev 2 r The magnetic field B at the center of the circular orbit is: B = ₀ I 2r = ₀ e v 4 r^2 Thus, B v r^2 . Substituting r 1 v^2 into the expression for B : B v ( 1 v^2 )^2 = v v^4 = v^5 Comparing this with B v^x , we get x = 5 . Answer: 5