Answer
a) $0.355\;\rm A$
b) $413\;\rm m/s$
Work Step by Step
a)
We know that the current is given by
$$I=\dfrac{Ne}{\Delta t}\tag 1$$
To find the current we need to find the total charge that passes some point in the ring during the periodic time.
We know that the magnitude of the particles' velocity around a circular path is given by
$$v=\dfrac{2\pi R}{\Delta t}$$
where $\Delta t$ here is the periodic time.
Solving for $\Delta t$;
$$\Delta t=\dfrac{2\pi R}{v}$$
Plugging into (1);
$$I=\dfrac{Ne}{\dfrac{2\pi R}{v}}=Ne\times\dfrac{v} {2\pi R}$$
And since the protons have a speed of $c$;
$$I= Ne\times\dfrac{c} {2\pi R}$$
Plugging the known;
$$I= 2\times 10^{14}\times 1.6\times 10^{-19}\times\dfrac{3\times 10^8} {2\pi \times 4.3\times 10^3}$$
$$I=\color{red}{\bf 0.355}\;\rm A$$
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b) We know that the kinetic energy of the car is given by
$$KE=\frac{1}{2}mv^2$$
So, its speed is given by
$$v=\sqrt{\dfrac{2KE}{m}}$$
The author told us that the car has kinetic energy as the protons beam.
$$KE=KE_{beam}=N\;KE_{\rm proton}$$
Thus;
$$v=\sqrt{\;\left(\dfrac{2 N\;KE_{\rm proton}}{m}\right)}$$
Plugging the known;
$$v=\sqrt{\dfrac{2 (2\times 10^{14})(4\times 10^{12})(2.6\times 10^{-19})}{1500}}$$
$$v=\color{red}{\bf 413}\;\rm m/s$$
