• # question_answer In the hydrogen atom the electron moves around the proton with a speed of$~2.0\times 106\text{ }m{{s}^{-1}}$in a circular orbit of radius $5.0\times {{10}^{11}}m.$What is the equivalent dipole moment? A) $2\times {{10}^{-24}}A{{m}^{2}}$                  B) $4\times {{10}^{-24}}A{{m}^{2}}$C) $8\times {{10}^{-24}}A{{m}^{2}}$                  D) $16\times {{10}^{-24}}A{{m}^{2}}$

The time period of electron motion in circular orbit is $T=\frac{2\pi a}{v}=\frac{2\pi \times 5.0\times {{10}^{-11}}}{2.0\times {{10}^{6}}}=5\pi \times {{10}^{-17}}\sec .$ Therefore, the equivalent current is $I=\frac{\text{Charge}}{\text{Time}}=\frac{e}{T}=\frac{1.6\times {{10}^{-19}}}{5\pi \times {{10}^{-17}}}=\frac{1.6}{5\pi }\times {{10}^{-2}}A$ Equivalent magnetic dipole moment = current$\times$area of circular orbit. $=I\times (\pi {{a}^{2}})=\frac{1.6\times {{10}^{-2}}}{5\pi }\times \pi {{(5\times {{10}^{-11}})}^{2}}=8\times {{10}^{-24}}\,A{{m}^{2}}.$ Hence, the correction option is [c].