Question

In a Bohr model of hydrogen atom, the electron circulates around the nucleus in the path of radius $5.1\times 10^{-11}m$ at a frequency of $6.8\times 10^{1}5rev/s$. Calculate the magnetic induction B at the centre of the orbit. What is the equivalent dipole moment ?

• A. $9.98\times 10^{-20}A{m}^2$
• B. $8.89\times 10^{-24}A{m}^2$
• C. $10\times 10^{-22}A{m}^2$
• D. $7.5\times 10^{10}A{m}^2$

Answer 1

Answer: (B)

Current flowing in the orbit. I = $\frac{e}{T}=e\nu$ $\left(\because \frac{I}{T}=\nu \right)$
Magnetic field at the centre of the current loop, B = $\frac{{\mu }_0}{4\pi }\frac{2\pi l}{r}=\frac{{\pi }_0}{4\pi }.\frac{2\pi e\nu }{r}$
= $\frac{10^{-7}\times 2\times 3.142\times 1.6\times 10^{-19}\times 6.8\times 10^{15}}{5.1\times 10^{-11}}$ = 13.41 T
Magnetic dipole moment of current loop is M = IA = $I\times \pi r^2=e\nu \times \pi r^2$
=