Question

An electron in orbit around a (one proton) nucleus has a speed of about 106 m/s. Deduce the equivalent current due to the motion of the electron and estimate the magnitude of the magnetic dipole moment of this one electron atom.

Answer 1

Answer:

(a) $-4.8\times 10^{-4}A$ (B) $4.22\times 106{-24}A/m^2$

Explanation:

We have given the the speed of the electron =10^6 m/sec

The time period of the electron is given by $T=\frac{2\pi r}{v}$ where r is the Bohr radius, which is equal to $5.29\times 10^{-11}m$

So time period $T=\frac{2\times \pi \times 5.29\times 10^{-11}}{10^6}=3.324\times 10^{-16}m/sec$

(A) We know that current $i=\frac{q}{T}=\frac{-e}{T}=\frac{-1.6\times 10^{-19}}{3.324\times 10^{-16}}=-4.8\times 10^{-4}A$

(B) Magnetic moment is given by $\mu =IA$ , where I is current and A is area

So $\mu =IA=-4.8\times 10^{-4}\times 3.14\times (5.29\times 10^{-11})^2=4.22\times 10^{-24}A/m^2$