When a particle with electric charge q moves with speed v in a plane perpendicular to a magnetic field B ,there is a magnetic force at right angles to the motion with magnetic qvB ,and the particle moves n a circle of radius r (see Figure 5.77). This equation for the magnetic force is correct even if the speed is comparable to the speed of light. Show that

$\mathbf{p}\mathbf{=}\frac{\mathbf{mv}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{(}\mathbf{\left|}\overrightarrow{\mathbf{v}}\mathbf{\right|}\mathbf{/}\mathbf{c}\mathbf{)}}^{\mathbf{2}}}}\mathbf{=}\mathbf{qBr}$even if is comparable to c.

This result is used to measure relativistic momentum: if the charge q is known, we can determine the momentum of a particle by observing the radius of a circular trajectory in a known magnetic field.

The data is given is that an electric charge is.

Speed is and perpendicular magnetic force is ${\mathrm{F}}_{\mathrm{m}}$and the equation is

${\mathrm{F}}_{\mathrm{m}}=\mathrm{qvB}$

And the charge travels at a speed near to that of light.

When the particle's speed approaches that of light, the relativistic momentum is measured. The magnetic field produces a net force that is equal to the rate of change in momentum.

${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{\mathbf{dt}}$

As a result, the net force matches the applied magnetic field, and the weight can be ignored. As a result, the magnetic force equals the rate of momentum change.

The magnetic field produces a net force that is equal to the rate of change in momentum.

$$\begin{gathered} F_{net}=\frac{d\overrightarrow{p}}{dt}................\left(1\right) \\ {F}_m=F_{net} \\ qvB=\frac{d\overrightarrow{p}}{dt}.................\left(2\right) \end{gathered}$$

The speed of the charge is given by the formula v , and the speed is defined as the change in distance over time.

$v=\frac{dr}{dt}$

So, to get will be inserted our expression into equation (2).

$$\begin{gathered} \mathrm{qvB}=\frac{\mathrm{d}\overrightarrow{\mathrm{p}}}{\mathrm{dt}} \\ \mathrm{qB}\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{\mathrm{d}\overrightarrow{\mathrm{p}}}{\mathrm{dt}} \\ \mathrm{qBdr}=\mathrm{dp} \end{gathered}$$

Integrate

$\mathrm{p}=\mathrm{qBr}............\left(3\right)$

At high speeds, the relativistic momentum technique will be used, in which the particle's momentum is given by

$p=\gamma mv$ ………………… (4)

Where mis the particle's mass, v is its speed, and $\gamma$is the relativistic factor that reflects the energy difference between motion and rest and is given by

$\gamma =\frac{1}{\sqrt{1-{(\left|\overrightarrow{v}\right|/c)}^2}}$

Whereis the speed of light, which is equal to$3\times {10}^{8}m/s$. As a result, in equation (4), the relativistic momentum will take the form

$p=\frac{1}{\sqrt{1-{(\left|\overrightarrow{v}\right|/c)}^2}}$ ………………… (5)

Because equations (3) and (5) have the same left side, the charged particle's momentum is the same.

$$\begin{gathered} p=qBr \\ =\frac{1}{\sqrt{1-{(\left|\overrightarrow{v}\right|/c)}^2}} \end{gathered}$$

So, the relation is proved.