Electronics and Communication Engineering - Materials and Components - Discussion
Discussion Forum : Materials and Components - Section 1 (Q.No. 34)
34.
If e is the charge of an electron, R is the radius of its orbit and ω is the angular velocity of electron, the magnetic dipole moment μm of the orbit is |μm| = eωR2.
Answer: Option
Explanation:
1 μm 1 = 0.5 eωr2.
Discussion:
6 comments Page 1 of 1.
Kavya said:
3 years ago
Anyone, please give a clear explanation.
Lekha said:
6 years ago
Revolution s about circle 360 degrees. 2pi = 6.28.
Mamon said:
8 years ago
How did you get the 6.28?
Is it a constant?
Is it a constant?
Siddharth said:
8 years ago
Magnetic dipole moment = Current x Area.
= Charge x Area / Time period of revolution.
Where time period of revolution is 6.28/Angular velocity,
= e x 3.14 R^2 x Angular velocity / 6.28.
= Charge x Area / Time period of revolution.
Where time period of revolution is 6.28/Angular velocity,
= e x 3.14 R^2 x Angular velocity / 6.28.
Priyanka said:
9 years ago
Please explain the solution.
(1)
Ravi mittal said:
10 years ago
Please explain?
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