Question

A circularly symmetrical magnetic field ( B depends only on the distance from the axis), pointing perpendicular to the page, occupies the shaded region in Fig. 5.58. If the total flux ($\mathbf{\int }\mathbf{}\mathbf{B}\mathbf{.}\mathbf{da}$) is zero, show that a charged particle that starts out at the center will emerge from the field region on a radial path (provided it escapes at all). On the reverse trajectory, a particle fired at the center from outside will hit its target (if it has sufficient energy), though it may follow a weird route getting there. [Hint: Calculate the total angular momentum acquired by the particle, using the Lorentz force law.]

Short answer

A charged particle that starts out at the center will emerge from the field region on a radial path, is proved.

Step-by-step solution

Significance of the magnetic field

The magnetic field is described as a region that is around a particular magnetic material or moving charge in which the magnetic force acts. The magnetic field is beneficial for distributing a magnetic force inside a magnetic material.

Determination of the momentum of a charged particle 

The equation of the angular momentum of a particle is expressed as:

$L=\int \frac{\mathrm{dL}}{\mathrm{dt}}\mathrm{dt}$

The above equation can also be reduced as:

$$\begin{gathered} \int \frac{dL}{dt}dt=\int Ndt \\ =\int \left(r\times F\right)dt \\ =\int \left(r\times q\right)(v\times B))dt \\ =\mathrm{q}\int r\times (dl\times B) \end{gathered}$$

Hence, further as:

$$\begin{gathered} \int \frac{dL}{dt}dt=q\int r\times (dl\times B) \\ =q\left[\int (r.B)dl-\int B(r.dl)\right] \end{gathered}$$

…(i)

As is mainly perpendicular to B , Hence,$r.B=0$.

The equation of the product of the distance and the increase in the length is expressed as:

$$\begin{gathered} r.dl=r.dr \\ =\frac{1}{2}d(r.r) \\ =\frac{1}{2}d{r}^2 \\ =rdr \end{gathered}$$

Hence, further as:

$r.dl=\left(\frac{1}{2\pi }\right)2\pi rdr$

Substitute the above value in equation (i).

$$\begin{gathered} L=\frac{q}{2\pi }{\int }_0^{R}B2\pi rdr \\ =-\frac{q}{2\pi }\int Bda \\ =-\frac{q}{2\pi }\Phi \end{gathered}$$

Hence, as$\mathrm{\Phi }=0$ , then the value of the angular momentum is L=0.

Thus, a charged particle that starts out at the center will emerge from the field region on a radial path, is proved.