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moves in a circular orbit of radius r about a nucleus. A uniform

magnetic field $$\vec B$$ is then established perpendicular to the plane of

the orbit. Assuming also that the radius of the orbit does not

change and that the change in the speed of the electron due to field $$\vec B$$ is small, find an expression for the change in the orbital magnetic

dipole moment of the electron due to the field.

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Solution

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Suppose the magnetic field increases linearly from zero to B in time t. According to Eq.

31-27, the magnitude of the electric field at the orbit is given by

$$E=\Bigg(\dfrac{r}{2}\Bigg)\dfrac{dB}{dt}=\Bigg(\dfrac{r}{2}\Bigg)\dfrac{B}{t}$$,

where r is the radius of the orbit. The induced electric field is tangent to the orbit and

changes the speed of the electron, the change in speed being given by

$$\Delta v=at=\dfrac{eE}{m_e}t=\Bigg(\dfrac{e}{m_e}\Bigg)\Bigg(\dfrac{r}{2}\Bigg)\Bigg(\dfrac{B}{t}\Bigg)t=\dfrac{erB}{2m_e}$$

The average current associated with the circulating electron is $$i = ev/2πr$$ and the dipole

moment is

$$\mu=Ai=(\pi r^2)\Bigg(\dfrac{ev}{2\pi r}\Bigg)\dfrac{1}{2}evr$$

The change in the dipole moment is

$$\mu=\dfrac{1}{2}er\Delta v =\dfrac{1}{2}er\Bigg(\dfrac{erB}{2m_e}\Bigg)=\Bigg(\dfrac{e^2r^2B}{4m_e}\Bigg)$$

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