Magnetics Ques 86
- In a hydrogen atom, the electron moves in an orbit of radius $0.5 \AA$ making $10^{16}$ revolutions per second. The magnetic moment associated with the orbital motion of the electron is ……
(1988, 2M)
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Answer:
Correct Answer: 86.$1.26 \times 10^{-23} A-m^{2}$
Solution:
Formula:
Magnetic Moment Of A Current Carrying Loop:
- Equivalent current $i=q f$
and magnetic moment $m=\left(i \pi r^{2}\right)=\pi q f r^{2}$
Substituting the values, we have
$ \begin{aligned} M & =(\pi)\left(0.5 \times 10^{-10}\right)^{2}\left(10^{16}\right)\left(1.6 \times 10^{-19}\right) \\ & =1.26 \times 10^{-23} A-m^{2} \end{aligned} $
BETA
