A proton moving in a magnetic field follows the curving path shown in Figure, traveling at constant speed in the direction shown. The dashed circle is the kissing circle tangent to the path when the proton is at location$\mathit{A}$. Refer to the directional arrows shown at the right in Figure when answering the questions below.

(a) When the proton is at location$\mathit{A}$, what is the direction of the proton's momentum?

(b) When the proton is at location $\mathit{A}$, what is the direction of$\mathit{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}\mathbf{/}\mathit{d}\mathit{t}$?

(c) The mass of a proton is$\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{}\mathit{k}\mathit{g}$. The proton is traveling at a constant speed of $\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$, and the radius of the kissing circle is $\mathbf{0}\mathbf{.}\mathbf{07}\mathbf{}\mathit{m}$. What is the magnitude of $\mathit{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}\mathbf{/}\mathit{d}\mathit{t}$of the proton?

A proton moving in a magnetic field follows the curving path shown in Figure. The dashed circle is the kissing circle tangent to the path when the proton is at location $A$. The proton is traveling at a constant speed of $\mathrm{v}=7.0\times {10}^5\mathrm{m}/\mathrm{s}$, and the radius of the kissing circle is $r=0.08\mathrm{m}$ . The mass of a proton is $m=1.7\times {10}^{-27}kg$.

According to the Newton's second law of motion, the net force is equal to the rate of change of momentum.

${\mathit{F}}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\frac{\mathbf{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}}{\mathbf{d}\mathbf{t}}$

Here,$\mathit{p}$is the momentum, and$\mathit{t}$is the time.

The rate of change of momentum$\left(\frac{d\stackrel{\rightharpoonup }{p}}{dt}\right)\hat{\mathbf{p}}$is non-zero for an object moving with a speed $\mathit{v}$ in a circular orbit, and it is equal to the $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$net$\mathbf{\parallel }$. And$\left|\stackrel{\rightharpoonup }{p}\right|\frac{\mathbf{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}}{\mathbf{d}\mathbf{t}}$is nonzero if direction is changing, and it is equal to$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$net $\mathbf{\perp }$. Write the expression for magnitude of momentum.

$\left|\stackrel{\rightharpoonup }{p}\right|\mathbf{=}\left|m\stackrel{\rightharpoonup }{v}\right|$

Here, $\left|\stackrel{\rightharpoonup }{v}\right|$ is the velocity of the proton, and$\mathit{m}$is the mass.

The net force on an object can be expressed as the sum of two parts and it is equal to the rate of change of the momentum. The two parts that we are taken about are the parallel rate of change of the momentum$\frac{d\stackrel{\rightharpoonup }{p}}{dt}$and the perpendicular rate of change of the momentum$\frac{d\stackrel{\rightharpoonup }{p}}{dt}$.

So, the change of momentum required is given by

$\frac{d\stackrel{\rightharpoonup }{p}}{dt}=\frac{d\stackrel{\rightharpoonup }{p}}{dt}+\frac{d\stackrel{\rightharpoonup }{p}}{d{t}_{\perp }}$

The parallel rate of change of the momentum changes the speed of the object and as we are given the speed is constant, therefore, the parallel rate is zero and it is equal to the rate change of the magnitude of the momentum

$$\begin{gathered} \frac{d\stackrel{\rightharpoonup }{p}}{dt}{\parallel }_{\parallel }=\frac{d\left|\stackrel{\rightharpoonup }{p}\right|}{dt}\hat{p} \\ \frac{d\stackrel{\rightharpoonup }{p}}{dt}{\parallel }_{\parallel }=0 \end{gathered}$$

This is the parallel direction of the momentum therefore, the momentum change in above equation will be due to the perpendicular force exerted on the proton and as shown in figure, at point $A$, the net force represented by the dashed line will be perpendicular on the momentum when the direction of the momentum is $\mathrm{d}$

The rate change here is the change in direction due to the perpendicular rate of change of the momentum. The magnitude of the perpendicular rate change equals the rate change of the direction of the momentum and at speeds much less than the speed of light is given by

$$\begin{gathered} \frac{d\stackrel{\rightharpoonup }{p}}{dt}{\perp }_{\perp }=\left|\stackrel{\rightharpoonup }{p}\right|\frac{d\stackrel{\rightharpoonup }{p}}{dt} \\ \frac{d\stackrel{\rightharpoonup }{p}}{dt}{\perp }_{\perp }=\frac{m{v}^2}{R} \end{gathered}$$

Where $m$is the mass of the proton. The direction of the rate change of the direction is toward the center of the kissing circle and it's perpendicular to the momentum $\stackrel{\rightharpoonup }{p}$. Hence, the direction will be as shown by the dashed line (head to the center) in figure, therefore, its direction represented by$b$

We are given the mass of the proton $m=1.7\times {10}^{-27}\mathrm{kg}$and the speed is constant with value$\mathrm{v}=6.0\times {10}^5\mathrm{m}/\mathrm{s}$ and the radius of the kissing circle is $R=0.07\mathrm{m}$.

Now we can plug our values for $m,v$and $R$into the above equation

$$\begin{gathered} d\stackrel{\rightharpoonup }{p}/dt=\frac{m{v}^2}{R} \\ =\frac{\left(1.7\times {10}^{-27}\mathrm{kg}\right){\left(6.0\times {10}^5\right)}^2}{0.07\mathrm{m}} \\ =8.75\times {10}^{-15}\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The magnitude of the proton is$8.75\times {10}^{-15}\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2$