Question

The magnetic field in a cyclotron that accelerates protons is $1.70T$. (a) How many times per second should the potential across the Dees reverse? (This is twice the frequency of the circulating protons.) (b) The maximum radius of the cyclotron is $0.250m$. What is the maximum speed of the proton? (c) Through what potential difference must the proton be accelerated from rest to give it the speed that you calculated in part (b)?

Short answer

The number of times potential across the Dees reverse per second is $v=5.18\times {10}^{7}Hz$, the maximum speed of the proton is $4.07\times {10}^{7}m/s$and the potential difference is $V=8.65\times {10}^{6}V$ must the proton be accelerated from rest to give it the speed.

Step-by-step solution

Definition of magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

Determine the number of times potential across the Dees reverse per second, the maximum speed of the proton and the potential difference must the proton be accelerated from rest to give it the speed.

The relation between angular frequency and magnetic field is $\omega =\frac{\left|q\right|B}{m}$

So

$\omega =\frac{\left(1.60\times {10}^{-19}C\right)\left(1.70T\right)}{1.67\times {10}^{-27}kg}$

$\omega =1.63\times {10}^{8}rad/s$

Than

$f=\frac{\omega }{2\pi }$

$f=\frac{1.63\times {10}^{8}rad/s}{2\pi }$

$f=2.59\times {10}^{7}Hz$

Since the number of times potential across the Dees reverse per second is twice of frequency $v=5.18\times {10}^{7}Hz$,

Hence, the number of times potential across the Dees reverse per second is

$v=5.18\times {10}^{7}Hz$

The kinetic energy is

$$\begin{gathered} K_{max}=\frac{\left(1.60\times {10}^{-19}C\right){\left(0.250m\right)}^2{\left(1.70T\right)}^2}{2\left(1.67\times {10}^{-27}kg\right)} \\ {K}_{max}=1.38\times {10}^{-12}J \\ {K}_{max}=8.653MeV \end{gathered}$$

The maximum speed is

$$\begin{gathered} v_{max}=\sqrt{\frac{2K_{max}}{m}} \\ {v}_{max}=\sqrt{\frac{2\left(1.38\times {10}^{-12}J\right)}{1.67\times {10}^{-27}kg}} \end{gathered}$$

$v_{max}=4.07\times {10}^{7}m/s$

Hence, the maximum speed of the proton is $4.07\times {10}^{7}m/s$and the potential difference is$V=8.65\times {10}^{6}V$must the proton be accelerated from rest to give it the speed.