Problem 56 - University Physics with Modern Physics

Question

The magnetic poles of a small cyclotron produce a magnetic field with magnitude 0.85 T. The poles have a radius of 0.40 m, which is the maximum radius of the orbits of the accelerated particles. (a) What is the maximum energy to which protons (\(q =\) 1.60 \(\times\) 10\(^{-19}\)C, \(m =\) 1.67 \(\times\) 10\(^{-27}\) kg) 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)? (d) For \(B =\) 0.85 T, what is the maximum energy to which alpha particles (\(q =\) 3.20 \(\times\) 10\(^{-19}\) C, \(m =\) 6.64 \(\times\) 10\(^{-27}\) kg) can be accelerated by this cyclotron? How does this compare to the maximum energy for protons?

Step-by-Step Solution

Answer

Maximum energy for protons: 2.89 MeV (joules to eV conversion). Time for one revolution: 150 ns. Magnetic field to double energy: 1.2 Tesla. Alpha particle energy: 1.44 MeV; less than proton energy.

Step 1: Determine Maximum Velocity

For a charged particle in a magnetic field, the maximum velocity is achieved when the centripetal force equals magnetic force. The formula for the force is \(qvB = \frac{mv^2}{r}\). Rearranging, we get \(v = \frac{qBr}{m}\). We'll use \(q = 1.60 \times 10^{-19}\, C\), \(B = 0.85\, T\), \(r = 0.40\, m\), and \(m = 1.67 \times 10^{-27}\, kg\) for protons. So, \(v = \frac{1.60 \times 10^{-19} \times 0.85 \times 0.40}{1.67 \times 10^{-27}}\).

Step 2: Calculate Kinetic Energy of Protons

With velocity \(v\) from Step 1, the kinetic energy \(K\) is given by \(K = \frac{1}{2}mv^2\). Substitute \(m = 1.67 \times 10^{-27}\, kg\) and \(v\) to calculate \(K\) in joules. Then convert this energy to electron volts using the conversion \(1\, J = 6.242 \times 10^{18}\, eV\).

Step 3: Calculate Time for One Revolution

The time \(T\) for one revolution is the circumference of the circle divided by the speed: \(T = \frac{2 \pi r}{v}\), where \(r = 0.40\, m\) and \(v\) is from Step 1.

Step 4: Calculate New Magnetic Field for Double Energy

To find the new magnetic field for double the energy, use the kinetic energy formula \(K = \frac{1}{2}mv^2\) and solve for \(B\) such that \(K\) becomes twice the previously found value. This requires \(B\) to be doubled if velocity is increased by a factor of \(\sqrt{2}\).

Step 5: Calculate Alpha Particle Energy

Repeat the earlier steps for alpha particles, with \(q = 3.20 \times 10^{-19}\, C\) and \(m = 6.64 \times 10^{-27}\, kg\). First calculate the velocity using \(v = \frac{qBr}{m}\), then compute the kinetic energy \(K\). Compare it with the proton kinetic energy found earlier.

Key Concepts

Magnetic Field Strength

The magnetic field strength is a crucial factor in a cyclotron, as it determines how effectively charged particles, such as protons, are accelerated. In the case of the provided exercise, a magnetic strength of 0.85 Tesla is used. This field strength creates centripetal force that makes charged particles move in a circular path.

The strength of the magnetic field, denoted by \(B\), impacts the maximum velocity that a particle can achieve. It's essential to ensure that the magnetic strength is optimal for the particles' intended acceleration, as it influences the radius of the particle's circular path inside the cyclotron.
The cyclotron uses strong electromagnets to produce this field, tightly controlling the paths of the charged particles. The tighter the magnetic field, the particle's trajectory can be more accurately confined, leading to higher energy gains. Thus, magnetic field strength is a cornerstone in effective cyclotron operation.

Proton Acceleration

Accelerating protons within a cyclotron involves utilizing the strong magnetic field to keep the particles traveling in a circular path. Protons are positively charged particles, with a charge \(q = 1.60 \times 10^{-19}\) C.

In the context of the cyclotron, protons move perpendicularly to the magnetic field lines, allowing the field to exert a magnetic force on them. This force keeps the protons moving in a circular trajectory, where the angle between the speed and magnetic field is always 90 degrees.

The formula used in calculating the maximum velocity of a proton is:

• \( v = \frac{qBr}{m} \)
• Here, \(q\) is the charge of the proton, \(B\) is the magnetic field strength, \(r\) is the maximum radius of the path, and \(m\) is the mass of the proton.

By determining the velocity, we can then calculate the kinetic energy, aiming to maximize the energy protons achieve within the cyclotron.

Alpha Particles

Alpha particles are another type of charged particle that can be accelerated using a cyclotron. An alpha particle consists of two protons and two neutrons, giving it a charge \(q = 3.20 \times 10^{-19}\) C and a mass \(m = 6.64 \times 10^{-27}\) kg.

Similar to protons, the maximum velocity for alpha particles is obtained using the same equation \( v = \frac{qBr}{m} \). However, due to their greater mass and charge compared to protons, alpha particles experience different dynamics in a magnetic field.

Calculating their energy involves the kinetic energy equation:

• \( K = \frac{1}{2}mv^2 \)

You can compare the maximum energy of alpha particles with protons to understand their relative energy dynamics. Due to their different physical properties, alpha particles require different magnetic field settings and adjustments to achieve desired levels of acceleration.

Centripetal Force

Centripetal force is a fundamental concept in understanding how particles are accelerated in a cyclotron. This force is what keeps particles moving in a circular path. In a cyclotron, when charged particles move perpendicular to a magnetic field, the magnetic force acts as the centripetal force necessary for circular motion.

The relationship between magnetic and centripetal force allows us to derive the formula for the velocity of a particle in a magnetic field:

• \( qvB = \frac{mv^2}{r} \)

From this, we rearrange to solve for velocity: \( v = \frac{qBr}{m} \). The centripetal force, thus, is crucial for achieving and maintaining the path and speed of particles in a cyclotron.

Understanding the balance of forces in play helps in calibrating a cyclotron's operation, ensuring effective acceleration, and optimizing the energy particles attain. This careful balance ensures that particles revolve correctly within the device, paving the way for successful experiments.