- #1
Javier Lopez
- 75
- 3
I tried to calculted the rotating speed fro a proton from the clasical point of view by using its magnetic moment and the moment of a rotating sphere uniformly charged as example here: https://ocw.mit.edu/courses/physics/8-07-electromagnetism-ii-fall-2012/exams/MIT8_07F12_quizsol2.pdf
It is said that the magnetic moment is:
$$\overrightarrow{m}=\frac{1}{5}Qwr^2 \overrightarrow{z}=0.4*\pi Qfr^2\overrightarrow{z}$$
By using the electric charge radius of a proton: 8.8783E-16 (0.8783 fm), its charge: 1.602176565E-19 coulombs and magnetic moment 2.59008941428763E-23 J/T (all SI), then I obtained a rotating frequency of 1.632e26 hz and an equatorial surface speed of 9.1e11 m/s that is over the c speed limit
Note: the maximum magnetic moment is with an equatorial point charge which magnetic momentum is: $$\overrightarrow{m}=\pi Qfr^2 \overrightarrow{z}$$
There is also the calculus of the residual magnetic field flux density:
$$m=\frac{Br}{\mu }*Volume=\frac{Br}{\mu }*\frac{4}{3 }\pi *r^3$$
Then I obtained Br=1.11e16 teslas
Is there something wrong?
It is said that the magnetic moment is:
$$\overrightarrow{m}=\frac{1}{5}Qwr^2 \overrightarrow{z}=0.4*\pi Qfr^2\overrightarrow{z}$$
By using the electric charge radius of a proton: 8.8783E-16 (0.8783 fm), its charge: 1.602176565E-19 coulombs and magnetic moment 2.59008941428763E-23 J/T (all SI), then I obtained a rotating frequency of 1.632e26 hz and an equatorial surface speed of 9.1e11 m/s that is over the c speed limit
Note: the maximum magnetic moment is with an equatorial point charge which magnetic momentum is: $$\overrightarrow{m}=\pi Qfr^2 \overrightarrow{z}$$
There is also the calculus of the residual magnetic field flux density:
$$m=\frac{Br}{\mu }*Volume=\frac{Br}{\mu }*\frac{4}{3 }\pi *r^3$$
Then I obtained Br=1.11e16 teslas
Is there something wrong?
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