91Ó°ÊÓ

• $B=0.100\text{ }T$
  

• The angle between velocity and the magnetic field is$\mathrm{Ï•}=89.0°$
• $K=2.00\text{ }keV$
  

We are given the angle between the velocity of the positron and its magnetic field. From this, we can find the velocity component perpendicular to the magnetic field. We have the formula for the radius of the orbital of the positron. We substitute the radius and velocity component in the formula for the time period, and we get the time period. We are also given the kinetic energy of the positron; from this, we can find the velocity ofthepositron. Once we get the velocity, we can find the pitch and the radius of its helical path.

Formula:

$T=\frac{2\mathrm{Ï€}r}{V}$

$p=v\cos \mathrm{ϕ}×T$

$r=\frac{m_{e}v\sin \mathrm{Ï•}}{eB}$

$K=\frac{1}{2}{m}_e{v}^2$

If$v$is the speed of the positron, then the speed of the positron along the plane perpendicular to the magnetic field will be, as the angle between velocity and field is given by.$\mathrm{Ï•}$

We know that:

$T=\frac{2\mathrm{Ï€}r}{v}$

We substitute the value for r andin the above equation:

$\begin{array}{c}T=\frac{2\mathrm{π}\left(\frac{m_{e}v\sin \mathrm{ϕ}}{eB}\right)}{v\sin \mathrm{ϕ}}\\ =\frac{2\mathrm{π}{m}_e}{eB}\\ =\frac{2\mathrm{π}(9.11×{10}^{−31})}{(1.60×{10}^{−19})(0.100)}\\ =3.58×{10}^{−10}\text{ s}\end{array}$

The time period for positron will be .$3.58×{10}^{−10}\text{ }s$

As we know, the kinetic energy for the positron:

$K=\frac{1}{2}{m}_e{v}^2$

$v^2=\frac{2K}{m_e}$

$v=\sqrt{\frac{2K}{m_e}}$

We have to convert kinetic energy into Joule as:

$\begin{array}{c}K=2.00×{10}^3\text{ }eV\\ =2.00×{10}^3\text{ }eV×\frac{1.60×{10}^{−19}\text{ }J}{1\text{ }eV}\\ =3.2×{10}^{−16}\text{ }J\end{array}$

$\begin{array}{l}v=\sqrt{\frac{2×3.2×{10}^{−16}}{9.11×{10}^{−31}}}\\ v=2.65×{10}^7\text{″¾/²õ}\end{array}$

We know that:

$\begin{array}{c}p=v\cos \mathrm{ϕ}×T\\ =2.65×{10}^7×cos89°×3.58×{10}^{−10}\\ =1.66×{10}^{−4}\text{ }m\end{array}$

The Pitch of the positron will be .$1.66×{10}^{−4}\text{ }m$

We know that:

$\begin{array}{c}r=\frac{m_{e}v\sin \mathrm{ϕ}}{eB}\\ =\frac{9.11×{10}^{−31}×2.65×{10}^7×sin89°}{1.60×{10}^{−19}×0.100}\\ =1.51×{10}^{−3}\text{ }m\end{array}$

The radius of the helical pathwill be.$1.51×{10}^{−3}\text{ }m$