Question

In a Bohr model of hydrogen atom, the electron circulates around the nucleus in the path of radius $5.1\times10^{-11}\,m$ at a frequency of $6.8\times10^15\,rev/s$. Calculate the magnetic induction B at the centre of the orbit. What is the equivalent dipole moment ?

• A. $9.98\times10^{-20}\,Am^2$
• B. $8.89\times10^{-24}\,Am^2$
• C. $10\times10^{-22}\,Am^2$
• D. $7.5\times10^{10}\,Am^2$

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 = $I \times \pi r^2 = e\nu \times \pi r^2$
=