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509. Problem 37.14 (RHK) An electron with kinetic energy
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travels in a circular path that is perpendicular to a uniform magnetic field,
subject only to the force of the field. We have to show (a) that the
magnetic dipole moment due to its orbital motion has magnitude
and that it is in the direction opposite to that of
.
(b) We have to find the magnetic dipole moment of a positive ion with kinetic
energy
under
the same circumstances. (c) An ionised gas consists of
electrons per
and
the same number of ions per
and the average ion kinetic energy to be
. We
have to calculate the magnetization of the gas for magnetic field of 1.18 T.
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Solution: Click For PDF Version (a)
Let the magnetic field be in the positive z-direction, which is perpendicular to the plane of the page and is coming out of it.
The speed of an electron of an electron of mass m in terms of its kinetic
energy
As the charge of electrons is
And
Under the action of the magnetic field
and
A uniformly circulating charge e is effectively a current
Current flowing in a closed loop enclosing area A is equivalent to a magnetic
dipole of moment
As the electron has negative charge and is moving in counter-clockwise
direction, the current flow will be in the clockwise direction, and therefore
the direction of the orbital magnetic dipole moment will be
We note that the orbital magnetic dipole moment of an electron in a magnetic
field
(b) For the case of a positive ion of kinetic energy
(c) We have to find the magnetization of the gas consisting of electrons and ions of equal number density
We are given that
and
The magnetic field is
Therefore, the magnetization of the gas, assuming that all electrons and ions are moving in circular paths perpendicular to an external magnetic field, which is net dipole moment per unit volume, will be
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.
and the
Lorentz force in a magnetic field
will be
.
.
That is 
.
.
,
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