Question

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.

Short answer

The relation $\mathrm{p}=\frac{\mathrm{mv}}{\sqrt{1-{\left(\left|\overrightarrow{\mathrm{v}}\right|\mathrm{c}\right)}^2}}$is proved.

Step-by-step solution

The given data

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.

The magnetic force equals the rate of momentum change

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.

Evaluate the magnetic force equals the rate of momentum

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

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

The momentum technique at high speed

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.