Question

Ratio of magnetic dipole moment to angular momentum of electron in first orbit

Answer 1

In classical physics, the magnetic field of a dipole is calculated as the limit of either a current loop or a pair of charges as the source shrinks to a point while keeping the magnetic moment m constant. For the current loop, this limit is most easily derived for the vector potential. Outside of the source region, this potential is (in SI units)[2]

{\displaystyle {\mathbf {A} }({\mathbf {r} })={\frac {\mu _{0}}{4\pi r^{2}}}{\frac {{\mathbf {m} }\times {\mathbf {r} }}{r}}={\frac {\mu _{0}}{4\pi }}{\frac {{\mathbf {m} }\times {\mathbf {r} }}{r^{3}}},} {\mathbf {A} }({\mathbf {r} })={\frac {\mu _{0}}{4\pi r^{2}}}{\frac {{\mathbf {m} }\times {\mathbf {r} }}{r}}={\frac {\mu _{0}}{4\pi }}{\frac {{\mathbf {m} }\times {\mathbf {r} }}{r^{3}}},

with 4π r2 being the surface of a sphere of radius r;

and the magnetic flux density (strength of the B-field) in teslas is[2]

Answer 2

Answer:

refer to the attachment. ....