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THE MAGNETIC FIELD. SLIDES BY ZIL E HUMA. OBJECTIVES. CIRCULATING CHARGES CYCLOTRON HALL EFFECT. CIRCULATING CHARGES. v. Electrons circulating in a chamber containing a gas at low pressure. The beam is made visible by collusions with the atoms of the gas. B. F B. G.

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The magnetic field

THE MAGNETIC FIELD

SLIDES BY ZIL E HUMA


Objectives
OBJECTIVES

  • CIRCULATING CHARGES

  • CYCLOTRON

  • HALL EFFECT


Circulating charges
CIRCULATING CHARGES

v

  • Electrons circulating in a chamber containing a gas at low pressure. The beam is made visible by collusions with the atoms of the gas.

B

FB

G

Uniform magnetic field B pointing out of the plane fills the chamber. The magnetic force FB is directed radially inward.





Thus the radius of the path is determined by force can be written as

1) the momentum p of the particles,

  • their charge,

  • and the strength of the magnetic field

    If the source of the electrons in above figure had projected them with a smaller speed, they would have moved in a circle of smaller radius.


The angular velocity of the circular motion is force can be written as

w= v/r

= |q|B/m

And the corresponding frequency is

v=w/2  =|q|B. /2m



The frequency given in above Eq. is called the cyclotron frequency, because particles circulate at this frequency in a cyclotron.


THE CYCLOTRON. frequency, because particles circulate at this frequency in a cyclotron.

The cyclotron is an accelerator that produces beams of energetic charged particles, which might be used in nuclear reaction experiments.


The final speed of the particles is determined by the radius R at which the particles leave the accelerator. Form Eq.

v =|q|BR/m

And the corresponding (non relativistic) kinetic energy of the particles is

K = 1/2 (mv2)

= q2B2R2 / 2m


THE HALL EFFECT. radius R at which the particles leave the accelerator. Form Eq.

In 1879 Edwin H. Hall conducted an experiment that permitted direct measurement of the sign and the number density (number per unit volume) of charge carriers in a conductor.

The Hall effect plays a critical role in our understanding of electrical conduction in metals and semi-conductors.


Consider a flat strip of material of width w carrying a current i.

The direction of the current is the conventional one.

A uniform magnetic field be is established perpendicular to the plane of the strip, such as by placing the strip between the poles of an electromagnet.

The charge carriers experience a magnetic deflecting force F.

F=qv * B


The charge carriers move to the right side of the strip. current i.

Positive charges moving in the direction of i experience a deflecting force in the same direction.

The build up of charge along the right side of the strip (and a corresponding deficiency of charge of that sign on the opposite side of the strip), which is the Hall Effect, produces an electric field E across the strip.


HALL POTENTIAL DIFFERENCE. current i.

A potential difference V=E/w, called the Hall Potential difference or Hall voltage exists across the strip.

We can measure by V by contacting the leads of the volt meter to points x and y.

The sign of V gives the sign of the charge carriers.

The magnitude of V gives their density (number per unit volume).


If the charge carriers are electron. current i.

For example, an excess of negative charge builds up on the right side of the strip and point y is at a lower potential than point x.

This may seem like an obvious conclusion in the case of mettle.


Let us assume that conduction in the material is due to charge carrier of a particular sign moving with drift velocity vd.

As the charge carrier drift, they are deflected to the right by the magnetic force, and they set up an electric field that acts inside the conductor to oppose the side ways motion of additional charge carriers.

Eventually an equilibrium is reached and the Hall voltage reaches its maximum.


The side ways magnetic force charge carrier of a particular sign moving with drift velocity vqvd * B is than balanced by the sideways electric force (qE).

qE+ qvd * B=0

E = - vd * B

Since vd and B are at right angles, so in terms of magnitudes

E= vd B


v charge carrier of a particular sign moving with drift velocity vd =j/ne

where j is the current density in the strip.

n is the density of charge carriers.

The current density j is the current i per unit cross-sectional area A of the strip. It t is the thickness of the strip then its cross-sectional area A =wt.

Substituting V/w for the electric field E, we obtain


V/w=v charge carrier of a particular sign moving with drift velocity vd B

=(j/ne)B

=(i/wtne)B

Solving for the density of the charge carriers

n = iB/etV


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