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Fundamentals Of Physics
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Short Answer

An electron with kinetic energy Ke travels in a circular path that is perpendicular to a uniform magnetic field, which is in the positive direction of a z axis. The electron’s motion is subject only to the force due to the field.

(a) Show that the magnetic dipole moment of the electron due to its orbital motion has magnitude μ=KeB B and that it is in the direction opposite that of B.(b)What is the magnitude of the magnetic dipole moment of a positive ion with kinetic energy role="math" localid="1662961253198" Ki under the same circumstances? (c) What is the direction of the magnetic dipole moment of a positive ion with kinetic energy Ki under the same circumstances? (d) An ionized gas consists of 5.3 ×1021electron/m3 and the same number density of ions. Take the average electron kinetic energy to be 6.2×10-20 J and the average ion kinetic energy to be 7.6 ×10-21J. Calculate the magnetization of the gas when it is in a magnetic field of 1.2 T.

  1. The magnitude of magnetic dipole moment of electron is μe=KeB, and its direction is along negative Z axis.
  2. The magnitude of magnetic dipole moment of positive ion is,μ+=K+B.
  3. The direction of magnetic dipole moment of positive ion is along the negative Z axis.
  4. Magnetization of ionized gas,M=3.1×102A/m
See the step by step solution

Step by Step Solution

Step 1: Listing the given quantities

Positive ion concentration, n+=5.3×1021 ions/m3

Electron concentration, ne=5.3×1021 electrons/m3

Average kinetic energy of electron, Ke=6.2×1020J

Average kinetic energy of positive ion, K+ =7.6×1021J

Applied magnetic field, B=(1.2 T)z^

Step 2: Understanding the concepts of magnetic moment

We will use the definition of magnetic moment and expression of current due to circular motion of electron and positive ion to find the magnetic moment of electron and positive ion. By using the necessary condition for circular motion, we can deduce the required expression.

Formula:

Force acting on charged particle moving in magnetic field is F=q(v×B)

Centripetal force, F=mv2rr^

Magnetic moment,μ=iA

K.E of mass m moving with velocity , K=12mv2

Step 3: (a) Calculations of the magnetic dipole moment of electron 

Magnetic moment of electron

μe=iA=eTA.

The electron is moving in a circular orbit of radius r. Hence the area vector is perpendicular to the plane of the orbit. We assume that the electron is moving in counter clockwise direction; then we get the direction of A as given below:

A=πr2z^

Using this, we get

Magnetic moment of electron

μe=eTA=eTπr2z^=eTπr2(z^)

Let’s assume that the electron is completing one revolution in time T=2πrv

Where, v is the velocity of electron.

Therefore,

μe=eTπr2(z^)=eπr2v2πr(z^)=evr2(z^)

Now the force acing on an electron moving in a magnetic field is F=e(v×B)

F=evB(r^)

As the electron is moving in circular orbit

F=evB(r^)=mev2r(r^)

This gives the magnitude of force as given below:

evB=mev2r

Therefore,

r=meveB

Substituting r in μe, we get

μe=ev2meveB(z^)=mev22B(z^)=KeB(z^)

Hence, the magnitude of magnetic dipole moment of electron is μe=KeB, and its direction is along the negative Z axis.

Step 4: (b) Calculations of the magnitude of magnetic dipole moment of a positive ion

Following similar steps for the positive ion, we will assume a charge on positive ion to be q=+e.

Magnetic moment of (+)ion is μ+ =iA=+eTA=+eTπr2(z^).

Let’s assume (+)ion is complete one revolution in time T=2πrv

where v is the velocity of (+)ion.

Therefore,

μ+ =+eTπr2(z^)=eπr2v2πr(z^)=evr2(z^)

Let F=evB(r^)

As the (+) ion is moving in circular orbit

r=m+veB

Substituting r in μ+. we get

μ+=ev2m+veB(z^)=m+v22B(z^)=K+B(z^)

Hence, the magnitude of magnetic dipole moment of a positive ion is localid="1662963848111" μ+=K+B.

Step 5: (c) Calculations of the direction of magnetic dipole moment of positive ion 

We have

μ+=K+B(z^)

Therefore, the direction of magnetic dipole moment of the positive ion is along the negative Z axis.

Step 6: (d) Calculations of magnetization of ionized gas

Magnetization of ionized gas:

Since the ionized gas is a mixture of positive ions and electrons, magnetization is

M=neμe+n+μ+=neKeB+n+K+B

Also ne=n+

Using this, we get

M=neB(Ke+K+)=5.3×10211.2(6.2×1020+7.6×1021)=5.3×1021×1021×(62+7.6)1.2=307.43.1×102A/m

Magnetization of ionized gas,M=3.1×102A/m

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