Question:

If the angular momentum of an electron revolving in a circular orbit is \(2.1\times10^{-34}\text{ Js}\), then the magnetic moment associated with the electron is (Specific charge of electron \(=1.76\times10^{11}\text{ C kg}^{-1}\)):

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Always remember the relation \(\mu=\frac{e}{2m}L\) between orbital magnetic moment and angular momentum.
Updated On: Jun 12, 2026
  • \(5.544\times10^{-23}\text{ Am}^2\)
  • \(0.924\times10^{-23}\text{ Am}^2\)
  • \(1.848\times10^{-23}\text{ Am}^2\)
  • \(3.696\times10^{-23}\text{ Am}^2\)
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The Correct Option is C

Solution and Explanation

Concept: Magnetic moment and angular momentum are related by \[ \mu=\frac{e}{2m}L \] Since \[ \frac{e}{m}=1.76\times10^{11}\text{ C kg}^{-1} \] therefore \[ \frac{e}{2m}=0.88\times10^{11} \]

Step 1:
Substitute the given values. \[ L=2.1\times10^{-34}\text{ Js} \] Hence, \[ \mu = 0.88\times10^{11} \times 2.1\times10^{-34} \] \[ = 1.848\times10^{-23}\text{ Am}^2 \] \[ \boxed{1.848\times10^{-23}\text{ Am}^2} \]
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