Subject: Applied Physics 2

Topic: Motion Of Charged Particles In Electric And Magnetic Fields

Difficulty: High

If a charge moving with velocity v enters into a region of a magnetic strength B , it experiences a Lorentz force. FL = e( v ⃗ x B ⃗ )

FL = e v B sinϴ

1.If v = 0 then $F_L$ = 0.that means magnetic force does not acts on a static charge.

2.If a charged particle enters the field parallel or anti parallel, force will be zero, It means field does not act on a charged particle moving along lines of induction.

1. Charged particle experiences maximum force when it enters perpendicular to the field.

Assume magnetic field acting into the page. when electron enters the perpendicular field, force is exerted on it, direction of which is given by Fleming left hand rule. This force will continuously deviate the path of electron and it will move along a curvilinear path. As the force does not act in the direction of v, magnitude of v is not changed. The only effect of $F_L$ is to change direction of motion. $_F$L is acting to the center so it is centripetal force . $F_L$ =m$v^2$/R

The centripetal force necessary to maintain the circular motion is provided by magnetic force.

m$v^2$ / R = Bev

Therefore R = mv/Be ……………………….(1)

Since all the parameters on RHS are constant, R is constant. The locus of points located at constant distance from a point is a circle. The electron describs a circular path. The time taken for one revolution T=(2П R)/v,

From equation (1) substituting value of R

T = 2Пm/Be……………………………….(2)

From equation 1 and 2 slower particle moves in a smaller circle while faster particles moves in a larger circle but all of them take same time for completing one revolution.

Deflection produced by magnetic field.

Let us assume restricted magnetic field confined to the region of length ‘l’ acting into the page. Motion of electron perpendicular to it. the electron entering the field bends through an arc and emerges from the field bends through an arc and emerges from the field and continues to travel tangentially afterwards. It strikes the screen at Q. PQ is the deflection caused by the field. Let ϴ be the angle between initial and final direction of electron.

tanϴ =Y/D--------------------------(1)

AB and CB are the tangent to arc AC at A and C respectively. Also

tanϴ =AC/R----------------------------(2)

for very small angle,

AC = l Y/D = l/R

Y = D l Be /mv

1/2 m$v^2$ = e $V_A$

Therefore, Y =D l B$\sqrt{(e/(2mVA )}$