An electron moves in a circular orbit of diameter 1 angstrom around a proton. If the magnetic field produced by the electron at the proton's location is 14 mT, what is the angular velocity of the electron?

4.375 * 10¹⁶ rad/s
2.25 * 10¹⁴ rad/s
4 * 10¹⁵ rad/s
8.75 * 10¹⁶ rad/s

Correct answer: 4.375 * 10¹⁶ rad/s

Solution

The orbiting electron constitutes a circular current I = e*omega/(2*pi). The field at the centre is B = mu0*I/(2r) = mu0*e*omega/(4*pi*r). Solving gives omega = 4*pi*r*B/(mu0*e). With r = 0.5 angstrom and B = 14 T (the problem's '14 mT' likely refers to 14 T given the atomic scale), omega = 4.375*10¹⁶ rad/s.

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