91Ó°ÊÓ

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 $7.0×{10}^{5}m/s$, and the radius of the kissing circle is 0.08 m . The mass of a proton is $1.7×{10}^{-27}kg$.

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

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

Here, pis the momentum, and tis the time.

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

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

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

(a)

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 $\frac{d\stackrel{→}{p}}{dt}$and the perpendicular rate of change of the momentum $\frac{d\stackrel{→}{p}}{dt}$. So, the change of the momentum required is given by

$\frac{d\stackrel{→}{p}}{dt}=\frac{d\stackrel{→}{p}}{dt}{âˆ\yen }_{âˆ\yen }+\frac{d\stackrel{→}{p}}{dt}âŠ\yen$

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,

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

So, the direction the rate change of the magnitude of the momentum is zero and it represents by

j direction from the figure.

(b)

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,

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

Where m , is the mass of the proton. Now we can plug our values for m,v and R into the above equation to get $\left|\stackrel{→}{p}\right|\frac{d\hat{p}}{dt}$,

$\begin{array}{rcl}\frac{m\stackrel{→}{v}}{R}& =& \left|\stackrel{→}{p}\right|\frac{d\hat{p}}{dt}\\ & =& \frac{\left(1.7×{10}^{-27}kg\right){\left(7×{10}^{5}m/s\right)}^2}{0.08m}\\ & =& 1.04×{10}^{-14}kg.m/s^2\end{array}$

The direction of the rate change of the direction is toward the center of the kissing circle and it's perpendicular to the momentum $\stackrel{→}{p}$. Hence, the direction will be as shown by the dashed line in figure, therefore, its direction represented by h .