8.2 Motion of a charged particle in an external magnetic field from Office of Academic Technologies on Vimeo.

**8.2 Motion of a charged particle in an external magnetic field**

Let’s consider that we have a uniform magnetic field pointing into the plane, and we send a positively charged particle with a velocity of **v** into this region such that the velocity of the charged particle is perpendicular to the magnetic field. As soon as this charged particle enters into this region, it will immediately feel the force generated by this external magnetic field, the magnetic force **F****B**, which is going to be equal to *q***v** cross **B**.

If we look at the direction of this force by applying right hand rule, holding the right hand fingers in the direction of the *q***v** vector, first vector, and then adjusting them and curling them towards the second vector, and that is pointing into the plane. We will see that the right hand thumb, which is the direction of the resultant vector, will be pointing in an upward direction, as perpendicular to both velocity vector and the magnetic field vector, which is pointing into the plane, in other words already perpendicular to this plane. So **F****B** will be perpendicular to both **v** and **B**, and pointing in an upward direction.

Now, if you visualize that, we have a particle moving with a velocity **v** to the right, and as soon as it enters into this region the forces pulling it in an upward direction under the influence of this force, then the particle will deviate from it’s original trajectory, and it is going to rotate like this. Whenever it comes to this point here, it’s velocity will be tangent to the trajectory, pointing like this, and in this position the force exerted by the external magnetic field from *q***v** cross **B** is going to be pointing to the left.

Now, under the influence of this force, therefore, it is going to continue to rotate, and let’s say come to this point. Now it’s velocity vector, again, tangent to the trajectory pointing to the left, and *q***v** cross **B** force is going to be pointing in the downward direction, which will cause even more rotation and the particle will rotate, and let’s say it will come to this point, the velocity vector in this position is pointing in downward direction and *q***v** cross **B** force will be pointing to the right.

We can see now the trajectory of the particle, how it will look like under the influence of the magnetic force generated by the external magnetic field. The particle is going to go along the circular trajectory. In other words, it’s going to end up by making circular motion.

Well, the magnitude of this force, is the magnitude of the first vector, *qv*, times the second vector which is *B* magnitude and times sine of the angle between these two vectors. Well, *q***v** is on the plane, and **B** is into the plane, therefore that angle is 90 degrees. So since sine 90 is just 1, then *F**B* is going to be equal to *qv* times *B*.

Well, since the particle is undergoing though a circular motion, and we know that for an object to make a circular motion, it has to be under the influence of a specific force, which is called “centripetal force”. So here in this case the necessary centripetal force is generated by the magnetic force. Again, as you recall from the circular motion, the direction of centripetal force is always towards the center of the trajectory. In other words it is always towards to center of the circle.

So since the cause of the circular motion is centripetal force, and this force is generated by the magnetic force, then *F**centripetal* should be equal to *F**B*. If we recall the centripetal force, that is defined as mass of the particle times *v*2 over *r*, where *r* is the radius of the trajectory, and *v*2 is the square of the magnitude of its tangential velocity. That should be equal to the magnitude of the magnetic force generated by this external magnetic field, which is *qvB*, and here we can cancel this *v*2, with the *v* on the right hand side, and solving for the radius of the trajectory we will end up with *mv* over *qB*.

Therefore, the radius of the circulating charge is going to be equal to this quantity. Here we can easily see that that radius is directly proportional to the velocity of the moving charge and as well as its mass. So here *m* is the mass of the charge, and *v *is the speed of the charge, and of course *q* is the charge of the charged particle and *B* is the external magnetic field.

So the radius of the orbit of the circulating charge is going to be directly proportional to the mass, and as well as the speed of the particle. The faster the particle will generate larger the radius. On the other hand, it is inversely proportional to the charge, so the greater the charge magnitude is going to result with the smaller the radius is, and also it is inversely proportional to the strength of the magnetic field. The weaker the magnetic field will result in the larger the radius.

Alright, when you were studying the circular motion, you have defined the angular velocity, which is basically the change in angular position with the respect to time, and this velocity was related to the tangential speed as *ω* is equal to *v* over *r*. So here *ω* is the angular velocity, or sometimes it is referred as also “angular frequency”.

If we use this expression and substitute *r* in terms of speed, mass of the charge and charge and the external magnetic field as *mv* over *qB*, the velocities will cancel, and we’re going to end up with *ω* is equal to *qB* over *m*. So the angular velocity, or the angular frequency of the circulating charge, therefore, is going to be equal to this quantity, angular frequency or the velocity of the charge.

Well, again, as you recall from the circular motion, the angular frequency is related to the linear frequency through an expression such that *ω* is equal to 2π times the linear frequency *f*. So here *f* represents the linear frequency. As you recall, frequency is the number of rotations per unit of time that an object is making in a circular motion, and duration of time for one full rotation is defined as the period. The relationship between the frequency and the period is such that frequency is equals to inverse of the period, and vice versa. So from here then we can express *ω* also as 2π divided by the period.

Now, if we solve this equation for the linear frequency, that will be equal to *ω* over 2π, or in explicit form this is going to be equal to 1 over 2π times *qB* over *m*. Of course the period is going to be just the opposite of this, and that will be 2π times *m* over *qB*. So these quantities are another important feature associated with the motion of a point charge which is making a circular motion in an external magnetic field under the influence of the magnetic force exerted by this field.

The frequency expression over here is also known as cyclotron frequency, and if the circulating charge is electron, then it is referred as electron cyclotron frequency. And if the circulating charge is an ion, then it is also in that case referred as ion cyclotron frequency. Well, if the charge enters into the external magnetic field region such that its velocity is perpendicular to the magnetic field, then we have seen that it goes along a circular trajectory like this.

Let’s look at a more general case such that the moving charge enters into the magnetic field region at an arbitrary angle between its velocity vector and the magnetic field vector, in other words, different than 90 degrees. For angle