0.7 Motion of a charged particle in magnetic field  (Page 3/5)

$$R=\frac{v}{\alpha B};T=\frac{2\pi }{\alpha B};\nu =\frac{\alpha B}{2\pi };\omega =\alpha B$$

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 :

$$\phi =\frac{2\pi }{T}Xt=\frac{2\pi qBt}{2\pi m}=\frac{qBt}{m}$$

Alternatively,

$$\phi =\omega t=\frac{qBt}{m}$$

Equations of motion

We consider circular motion of a particle carrying a positive charge q moving in x-direction with velocity ${\mathbf{v}}_0$ 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,

$${\mathbf{v}}_0=v_0\mathbf{i};\mathbf{B}=-B\mathbf{k}$$

where $v_0$ 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.

Motion of particle carrying charge

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 :

$$\mathbf{v}=v_x\mathbf{i}+v_y\mathbf{j}$$

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 :

$$\Rightarrow \mathbf{v}=v_0\cos \phi \mathbf{i}+v_0\sin \phi \mathbf{j}$$

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

$$\phi =\omega t$$

Substituting this in the expression of velocity,

$$\Rightarrow \mathbf{v}=v_0\cos \omega t\mathbf{i}+v_0\sin \omega t\mathbf{j}$$

Again substituting for angular speed,

$$\Rightarrow \mathbf{v}=v_0\cos \frac{qBt}{m}\mathbf{i}+v_0\sin \frac{qBt}{m}\mathbf{j}$$

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 :

$$\mathbf{r}=x\mathbf{i}+y\mathbf{j}$$ $$\Rightarrow \mathbf{r}=R\sin \phi \mathbf{i}+\left(R-R\cos \phi \right)\mathbf{j}$$

Substituting for R and φ,

$$\Rightarrow \mathbf{r}=\frac{m{v}_0}{Bq}[\sin \frac{qBt}{m}\phi \mathbf{i}+\left(1-\cos \frac{qBt}{m}\right)\mathbf{j}]$$

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.

Motion of charged particle entering a magnetic field

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.