Problem 61 - University Physics with Modern Physics

Question

The magnetic poles of a small cyclotron produce a magnetic field with magnitude 0.85\(\mathrm { T }\) . The poles have a radius of \(0.40 \mathrm { m } ,\) which is the maximum radius of the orbits of the accelerated particles. (a) What is the maximum energy to which protons \(\left( q = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m = 1.67 \times 10 ^ { - 27 } \mathrm { kg } \right)\) can be accelerated by this cyclotron? Give your answer in electron volts and in joules. (b) What is the time for one revolution of a proton orbiting at this maximum radius? (c) What would the magnetic-field magnitude have to be for the maximum energy to which a proton can be accelerated to be twice that calculated in part (a)? For \(B = 0.85 \mathrm { T } ,\) what is the maximum energy to which alpha particles \(\left( q = 3.20 \times 10 ^ { - 19 } \mathrm { C } , m = 6.65 \times 10 ^ { - 27 } \mathrm { kg } \right)\) can be accelerated by this cyclotron? How does this compare to the maximum energy for protons?

Step-by-Step Solution

Answer

The max energy for protons is about 5.06 × 10^-13 J or 3.16 MeV. One revolution takes about 4.92 × 10^-8 s. The magnetic field for double energy: 1.20 T. For alpha particles: max 1.01 × 10^-12 J or 6.34 MeV, twice protons' energy.

Step 1: Calculate kinetic energy of protons in joules

The kinetic energy of a particle in a cyclotron can be calculated using the formula \( K = \frac{1}{2} m v^2 \). The velocity \(v\) of the particle is related to the magnetic field \(B\), charge \(q\), and radius \(r\) by \( v = \frac{qBr}{m} \). By substituting this into the kinetic energy equation, we find \( K = \frac{1}{2} m \left( \frac{qBr}{m} \right)^2 = \frac{q^2 B^2 r^2}{2m} \). Substituting the values for protons: \( q = 1.60 \times 10^{-19} \text{ C} \), \( B = 0.85 \text{ T} \), \( r = 0.40 \text{ m} \), and \( m = 1.67 \times 10^{-27} \text{ kg} \), we get:\[ K = \frac{(1.60 \times 10^{-19})^2 \times (0.85)^2 \times (0.40)^2}{2 \times 1.67 \times 10^{-27}} \approx 5.06 \times 10^{-13} \text{ J} \].

Step 2: Convert joules to electron volts for protons

The energy in electron volts can be obtained by converting joules using the relation: \(1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}\). Thus, the energy in electron volts is:\[ K_{(eV)} = \frac{5.06 \times 10^{-13}}{1.60 \times 10^{-19}} \approx 3.16 \times 10^6 \text{ eV} \].

Step 3: Calculate time for one revolution of a proton

The cyclotron frequency or angular frequency for a proton is given by \( f = \frac{qB}{2\pi m} \). The period \( T \) for one revolution is the inverse of this frequency:\[ T = \frac{2\pi m}{qB} \]. Substituting the given values for protons, \( q = 1.60 \times 10^{-19} \text{ C} \), \( B = 0.85 \text{ T} \), and \( m = 1.67 \times 10^{-27} \text{ kg} \), we obtain:\[ T = \frac{2\pi \times 1.67 \times 10^{-27}}{1.60 \times 10^{-19} \times 0.85} \approx 4.92 \times 10^{-8} \text{ s} \].

Step 4: Calculate required magnetic field for twice the energy

To achieve twice the maximum energy calculated in Part (a), the following relationship must hold: \( K_{new} = 2K \). Therefore, doubling \( K \) implies two times the square of the original magnetic field as per the equation:\( \frac{B^2}{2m} = 2 \times \frac{B_{original}^2}{2m} \). Solving this for the new magnetic field \(B_{new}\):\[ B_{new} = \sqrt{2} \times B_{original} \approx \sqrt{2} \times 0.85 \text{ T} \approx 1.20 \text{ T} \].

Step 5: Calculate kinetic energy for alpha particles at given field

Following the same steps and using the relationship, which provides kinetic energy for alpha particles \( K_{\alpha} = \frac{q_{\alpha}^2 B^2 r^2}{2m_{\alpha}} \), substituting the values \( q_{\alpha} = 3.20 \times 10^{-19} \text{ C} \), \( B = 0.85 \text{ T} \), \( r = 0.40 \text{ m} \), and \( m_{\alpha} = 6.65 \times 10^{-27} \text{ kg} \), we calculate:\[ K_{\alpha} = \frac{(3.20 \times 10^{-19})^2 \times (0.85)^2 \times (0.40)^2}{2 \times 6.65 \times 10^{-27}} \approx 1.01 \times 10^{-12} \text{ J} \].Convert to electron volts: \[ K_{\alpha(eV)} = \frac{1.01 \times 10^{-12}}{1.60 \times 10^{-19}} \approx 6.34 \times 10^6 \text{ eV} \].

Key Concepts

Magnetic Field Strength

A cyclotron uses a magnetic field to accelerate charged particles, like protons and alpha particles. The magnetic field strength, denoted as \(B\), is crucial because it determines the radius of the particle's orbit and its maximum kinetic energy. In our cyclotron problem, the magnetic field has a magnitude of 0.85 Tesla (T). This strong magnetic field allows particles to maintain a circular path within the cyclotron.
The reason the magnetic field is so important is that it interacts with the charge of the proton, giving it centripetal force which keeps it in its orbit. Without the magnetic field, the particles would not be able to gain the necessary speed and energy needed in particle accelerators like the cyclotron. Increasing \(B\) increases the particle’s velocity, thereby raising the kinetic energy.

Proton Acceleration

In a cyclotron, protons are accelerated using both electrical and magnetic fields. The process starts with an electric field giving the proton an initial push. Once inside the cyclotron, the magnetic field takes over, causing the proton to follow a circular path. As the proton moves, it continuously crosses a gap between two magnetic regions called 'dees.'

• Each time it crosses the gap, the direction of the electric field changes, accelerating the proton.
• This timing is critical and must match the natural frequency of the proton's orbit. This ensures that the proton gains energy with each pass.

By the time it reaches the extremity of the cyclotron's maximum radius, it attains maximum kinetic energy, determined by the magnetic field strength and the radius of the cyclotron.

Particle Kinetic Energy

Kinetic energy of a particle in a cyclotron is expressed through the formula \( K = \frac{1}{2} mv^2 \), where \(m\) is mass and \(v\) is velocity. However, in cyclotron physics, velocity \(v\) can be expressed as \( \frac{qBr}{m} \), particularly due to the magnetic influence. By substituting this in, we get \( K = \frac{q^2 B^2 r^2}{2m} \).
For protons, using the given values, the maximum kinetic energy obtained calculates to approximately \(5.06 \times 10^{-13}\) Joules or \(3.16 \times 10^6\) electron volts.
In cyclotron settings, maximizing kinetic energy is key for subsequent physics experiments needing high-speed protons.

Revolution Time

The revolution time of a proton in a cyclotron tells us how quickly it circulates within the magnetic field. This time, represented by \(T\), is calculated using the formula \( T = \frac{2\pi m}{qB} \).

• This formula reveals that the time is directly proportional to the mass of the particle and inversely proportional to both its charge and the magnetic field strength.
• Using the given data for protons, the revolution time comes out to approximately \(4.92 \times 10^{-8}\) seconds.

This extremely short time indicates how rapidly a proton can circle the cyclotron, crucial for achieving high levels of acceleration.

Energy Conversion

Converting energy into different units allows for better interpretation and comparison across various physics applications. In our exercise, kinetic energy was initially obtained in Joules, then converted to electron volts (eV).

• 1 Joule equates to \(1.60 \times 10^{-19}\) eV, a conversion factor necessary due to the smaller scale of quantities dealt in particle physics.
• This conversion is particularly useful because one electron volt represents the kinetic energy gained by an electron when it is accelerated through an electrical potential difference of one Volt.

For protons, this translation shows the practical energy level they acquire, in terms much more common and intuitive in high-energy physics.

Alpha Particles

Alpha particles, differing from protons, consist of two protons and two neutrons, making them heavier and carrying more charge. In this exercise, they have a charge of \(3.20 \times 10^{-19}\) C and a mass of \(6.65 \times 10^{-27}\) kg.
The kinetic energy for alpha particles under the same magnetic field is calculated similarly to protons, where \( K_{\alpha} = \frac{q_{\alpha}^2 B^2 r^2}{2m_{\alpha}} \).
The result for the cyclotron indicates a kinetic energy of \(1.01 \times 10^{-12}\) Joules or \(6.34 \times 10^6\) eV.
Given these properties, alpha particles reach higher kinetic energy levels than protons, highlighting their potential for specific scientific experiments.