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# Motion of a charged particle in magnetic field

Module by: Sunil Kumar Singh. E-mail the author

Motion of a charged particle in magnetic field is characterized by the change in the direction of motion. It is expected also as magnetic field is capable of only changing direction of motion. In order to keep the context of study simplified, we assume magnetic field to be uniform. This assumption greatly simplifies the description and lets us easily visualize the motion of a charged particle in magnetic field.

Lorentz magnetic force law is the basic consideration here. Hence, we shall first take a look at the Lorentz magnetic force expression :

F = q v X B F = q v X B

1: There is no magnetic force on a stationary charge (v=0). As such, our study here refers to situations in which charge is moving with certain velocity in the magnetic field. This condition is met when the charge is released with certian velocity in the magnetic field.

2: The magnetic field (B) is an uniform stationary magnetic field for our consideration in the module. It means that the magnitude and direction of magnetic field do not change during motion. The charged particle, however, is subjected to magnetic force acting side way. The direction of motion of charged particle, therefore, changes. In turn, the direction of magnetic force being perpendicular to velocity also changes. Important point to underline here is that this loop of changing directions of velocity and magnetic force is continuous. In other words, the directions of both velocity and magnetic force keeps changing continuously with the progress of motion.

This aspect of continuous change is shown in the figure below. Note that direction of magnetic field is fixed in y-direction. Initially, the charged particle is at the origin of coordinate reference with a velocity v in x-direction. Applying right hand rule for vector cross product and considering a point positive charge, we see that magnetic force is directed in z-direction. As a result, the particle is drawn to move along a curved path with velocity (having same speed) directed tangential to it. The magnetic force vector also changes sign being perpendicular to velocity vector. In this manner, we see that the directions of both velocity and magnetic force keeps changing continuously as pointed out.

3: The nature of motion depends on the initial directions of both velocity and magnetic field. The initial angle between velocity and magnetic field ultimately determines the outcome i.e. nature of motion.

We shall, therefore, discuss motion of charged particular on the basis of the enclosed angle (θ) between velocity and magnetic field vectors. There are following three cases :

• The motion of the charged particle is along the direction of magnetic field.

• The motion of the charged particle is perpendicular to the direction of magnetic field.

• The motion of the charged particle is neither along nor perpendicular to the direction of magnetic field.

## Motion of the charged particle along magnetic field

There are two possibilities. The enclosed angle (θ) is either 0° or 180°. In either case, sine of the angle is zero. Therefore, magnetic force is zero and the motion of particle remains unaffected (of course here we assume that there is no other force field present).

## Motion of the charged particle perpendicular to magnetic field

This is the case in which charged particle experiences maximum magnetic force. It is given by :

F = q B v sin 90 ° = q v B F = q B v sin 90 ° = q v B

If the span of magnetic field is sufficient around the charged particle, then it will describe a circular path as magnetic force is always perpendicular to the motion. The magnetic force provides the centripetal force required for circular motion. The span of magnetic field around charged particle is important. Here, we consider some interesting cases as shown in the figure. For all cases, we assume that motion of charged particle is in the plane of the drawing and magnetic field is perpendicular and into the plane of drawing. Magnetic field is shown by evenly distributed X sign indicating that it is an uniform magnetic field directed into the plane of drawing.

In the first case, there is sufficient span of magnetic field around charged particle and it is able to describe circular path. In second case, the charged particle enters the region of magnetic field and never completes the circular trajectory. Similarly, the charged particle in third case also does not complete the circular path as it comes out of the region of magnetic field even before completing half circle.

Now, we consider the first case in which the charged particle is able to complete circular path. Let the mass of the particle carrying charge is m. Then, magnetic force is equal to centripetal force,

m v 2 R = q v B m v 2 R = q v B m v R = q B m v R = q B

The radius of circular path, R, is given as :

R = m v q B R = m v q B

We can easily interpret the effects of various parameters in determining the radius of circular path. Greater charge and magnetic field result in smaller radius. On the other hand, greater mass and speed result in greater radius. Now, the time period of revolution is :

T = 2 π R v = 2 π m q B T = 2 π R v = 2 π m q B

Frequency of revolution is :

ν = 1 T = q B 2 π m ν = 1 T = q B 2 π m

Angular speed is :

ω = 2 π ν = 2 π q B 2 π m = q B m ω = 2 π ν = 2 π q B 2 π m = q B m

Important aspect of these results is that properties related to periodicity of revolutions i.e. time period, frequency and angular velocity all are independent of the speed of the particle. It is a very important result which is used in cyclotron (to accelerate charged particle) to synchronize with the frequency of application of electric field. We shall learn about this in another module.

### Specific charge

The ratio of charge and mass of the particle is known as specific charge and is denoted by α. Evidently, its unit is Coulomb/kg. This quantity is important in describing motion of charged particle in magnetic field. We observe that magnetic force is proportional to charge q, whereas acceleration of the particle carrying charge is inversely proportional to mass m. Clearly, the effects of these two quantities are opposite and hence they appear as the ratio q/m in most of the formula describing motion. Recasting formulas with specific charge, we have :

R = v α B ; T = 2 π α B ; ν = α B 2 π ; ω = α B R = v α B ; T = 2 π α B ; ν = α B 2 π ; ω = α B

### Angular deviation

Having known the time period, it is easy to know the angle subtended at the center by the arc of travel during the motion in a particular time interval. Since time period T corresponds to a angular travel of 2π, the angular travel or deviation (φ) corresponding to any time travel, t, is :

φ = 2 π T X t = 2 π q B t 2 π m = q B t m φ = 2 π T X t = 2 π q B t 2 π m = q B t m

Alternatively,

φ = ω t = q B t m φ = ω t = q B t m

### Equations of motion

We consider circular motion of a particle carrying a positive charge q moving in x-direction with velocity v 0 v 0 in a uniform magnetic field B, which is perpendicular and into the plane of drawing. Let xy be the plane of drawing and -z be the direction of magnetic field. Here,

v 0 = v 0 i ; B = - B k v 0 = v 0 i ; B = - B k

where v 0 v 0 is the magnitude of velocity. Applying Right hand rule of vector cross product, we see that magnetic force F is directed in y-direction. These initial orientations are shown in the figure assuming that we begin our observation of motion when particle is at the origin.

The magnetic force F provides the necessary centripetal force for the particle to execute circular motion in xy plane in anticlockwise direction with center of circle lying on y-axis. Let the particle be at a point P after time t. Expressing velocity vector in components :

v = v x i + v y j v = v x i + v y j

Let the velocity vector makes an angle φ with the x-axis. As the magnitude of velocity does not change due to magnetic force, we have :

v = v 0 cos φ i + v 0 sin φ j v = v 0 cos φ i + v 0 sin φ j

Since particle is executing a uniform circular motion with a constant angular speed,

φ = ω t φ = ω t

Substituting this in the expression of velocity,

v = v 0 cos ω t i + v 0 sin ω t j v = v 0 cos ω t i + v 0 sin ω t j

Again substituting for angular speed,

v = v 0 cos q B t m i + v 0 sin q B t m j v = v 0 cos q B t m i + v 0 sin q B t m j

This is the expression of velocity at any time "t" after the start of motion. Let the displacement vector of the particle from the origin is r. Then :

r = x i + y j r = x i + y j r = R sin φ i + R R cos φ j r = R sin φ i + R R cos φ j

Substituting for R and φ,

r = m v 0 B q [ sin q B t m φ i + 1 cos q B t m j ] r = m v 0 B q [ sin q B t m φ i + 1 cos q B t m j ]

### Motion of charged particle entering a magnetic field

The charged particle entering a magnetic field describes an arc which is at most a semicircle. If the span of magnetic field is limited, then there is no further bending of path due to magnetic force. Let us consider a case in which a particle traveling in the plane of drawing enters a region of magnetic field at angle α.

We should realize here that even though the charged particle enters magnetic region obliquely (i.e at an angle) in the plane of motion, the directions of velocity and magnetic field are still perpendicular to each other. The particle, in turn, follows a circular path. However, the particle needs to move in the region behind the boundary YY’ in order to complete the circular path. But, there is no magnetic field behind the boundary. Therefore, the charged particle is unable to complete the circular path. From geometry, it is clear that point of entry and point of exit are points on the circle which is intersected by the boundary YY’. By symmetry, the angle that the velocity vector makes with the boundary YY’ at the point of entry is same as the angle that velocity vector makes with the boundary YY' at the point of exit.

By geometry, the angle between pair of lines is same as the angle between the lines perpendicular to them. Hence,

O A D = C O D = α O A D = C O D = α

and

A O C = 2 α A O C = 2 α

The length of arc, AEC is :

l = A E C = 2 α R l = A E C = 2 α R

Substituting for R, we have :

l = 2 α m v q B l = 2 α m v q B

The time of travel in the magnetic field is :

t = l v = 2 α m q B t = l v = 2 α m q B

When charged particle enters magnetic field at right angle, velocity vector is perpendicular to the boundary of magnetic field. We know that a tangent can be drawn on a circle in this direction only at the points obtained by the intersection of the circle by the boundary line which divides the circle in two equal sections. A charged particle can, therefore, travel a semicircular path when it enters into the region magnetic field at right angle, provided of course the span of magnetic is sufficient.

We should understand that circular arc path as obtained by the analysis above can be subject to availability of magnetic field till the charged particle begins to move backwards. For a smaller extent of the magnetic field, we find that the particle emerges out of the magnetic field without being further deviated. If the extent of magnetic field is greater than or equal to R, then charged particle describes up to a semicircle depending on the angle at which it enters magnetic region. However, if the extent of magnetic field is less than R, then particle emerges out of the magnetic field without being further deviated.

## Motion of the charged particle oblique to magnetic field

This is the general case of motion of a charged particle in magnetic field. Here, velocity and magnetic field vectors are at an acute angle θ. In order to study the motion, we resolve the velocity vector such that one of the components is parallel and other is perpendicular to the magnetic field.

v | | = v cos θ v | | = v cos θ v = v cos θ v = v cos θ

The velocity component perpendicular to the magnetic field results in a magnetic force which provides the necessary centripetal force for the particle to move along a circular path as discussed in previous section. On the other hand, the velocity component parallel to magnetic field results in zero magnetic force and motion in this direction is unaffected due to this component of velocity. The charged particle moves in this direction without being accelerated.

We can visualize superimposition of two motions. The initial conditions of set up are shown in the figure in which particle is shown to have velocity v at the origin of coordinate system. The magnetic field is directed in x-direction. The magnetic force (F) due to perpendicular component of velocity and magnetic field is directed in negative z-direction.

If we ignore the parallel component of velocity, then particle will follow circular path due to perpendicular component of velocity in y-z plane as shown here :

But, there is a component of velocity in x-direction. The charged particle still completes a revolution, but not in the circular plane because charged particle also moves in the direction perpendicular to the circular plane. The net result is that the path of revolution is a stretched out series of circles in the form of a helix.

The expression for radius is similar as that for the circular motion under magnetic field (earlier case). The only change is that v is exchanged by v v .

R = m v q B = m v sin θ q B R = m v q B = m v sin θ q B

The expressions for time period, frequency and angular velocity etc do not change as these parameters are independent of velocity.

The distance between two consecutive points in x-direction determines the pitch of the helical path. This distance in x-direction is traveled by the particle with the parallel component of velocity in the time in which particle completes a revolution. If T be the time period of revolution, then pitch, p, of the helical path is :

p = v | | T = v T cos θ = 2 π m v cos θ q B p = v | | T = v T cos θ = 2 π m v cos θ q B

### Example 1

Problem : An electron with a kinetic energy of 10 eV moves into a region of uniform magnetic field of 5 X 10 - 4 5 X 10 - 4 T. The initial angle between velocity and magnetic field vectors is 60 degree. Determine the pitch of resulting helical motion.

Solution : The expression of pitch of helical path is :

p = 2 π m v cos θ q B p = 2 π m v cos θ q B

We notice here that speed is not directly given. However, kinetic energy is given in electron volt unit. By definition, an electron volt is equal to kinetic energy gained by an electron while passing through a potential difference of 1 V. We get kinetic energy in Joule by multiplying electron-volt value by 1.6 X 10 - 19 1.6 X 10 - 19 .

K = m v 2 2 = 10 e V = 10 X 1.6 X 10 - 19 J = 16 X - 19 J K = m v 2 2 = 10 e V = 10 X 1.6 X 10 - 19 J = 16 X - 19 J v = 2 K m = 2 X 16 X 10 - 19 9.1 X 10 - 31 = 3.52 X 10 12 = 1.88 X 10 6 m / s v = 2 K m = 2 X 16 X 10 - 19 9.1 X 10 - 31 = 3.52 X 10 12 = 1.88 X 10 6 m / s

Putting values in the expression of pitch :

p = 2 π X 9.1 X 10 - 31 X 1.88 X 10 6 X 0.5 1.6 X 10 - 19 X 5 X 10 - 4 p = 2 π X 9.1 X 10 - 31 X 1.88 X 10 6 X 0.5 1.6 X 10 - 19 X 5 X 10 - 4 p = 6.71 X 10 - 2 m = 6.71 c m p = 6.71 X 10 - 2 m = 6.71 c m

## Magnetic bottle

In plasma research, one of the main tasks is to contain plasma (ions or charged elementary particles). Plasma particles can not be restrained in any material confinement because of extraordinarily high temperature associated with them. A magnetic bottle is an arrangement of two magnetic sources (solenoids or any other magnetic source) which produce magnetic fields. The arrangement is such that direction of magnetic field is from one solenoid to another. The magnetic field between two solenoids is non-uniform. It is stronger near the solenoid and weaker in the middle. See that lines of force are denser near the solenoids and rarer in the middle.

A charged particle is in the helical motion in this magnetic region. As it moves in stronger magnetic region near the solenoid, the radius of helical path is smaller. On the other hand, the radius of helical path is greater in the middle as magnetic field is weaker there.

R = m v q B R = m v q B

As the particle reaches towards the solenoid i.e. end of the arrangement, it is rebounded because there is a component of magnetic force pointing towards the central part of the arrangement. See figure that how force components point toward middle. This component decelerates the particle till it stops and starts moving in opposite direction. The stronger magnetic region near the solenoid, therefore, functions as reflector of charged particles.

In this manner, plasma particles are confined within a region due to suitably designed magnetic field. The whole arrangement works like a bottle for the charged particles and hence is called magnetic bottle.

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