Cyclotron Magnetic Field Calculation

This problem requires us to determine the strength of the magnetic field needed within a cyclotron to accelerate a deuteron. A cyclotron is a type of particle accelerator that accelerates charged particles using a constant magnetic field and an oscillating electric field.

Understanding Cyclotron Physics

In a cyclotron, a charged particle traverses a spiral path. The magnetic field, applied perpendicular to the dees (D-shaped electrodes), provides the necessary centripetal force to keep the particle moving in a circular path. The frequency of the alternating voltage applied across the gap between the dees is synchronized with the revolution frequency of the particle (the cyclotron frequency). This ensures that the particle is accelerated each time it crosses the gap.

• The magnetic force acting on the charged particle is given by $F_B = qvB$, where $q$ is the charge, $v$ is the velocity, and $B$ is the magnetic field strength.
• This magnetic force acts as the centripetal force, $F_C = \frac{mv^2}{r}$, where $m$ is the mass and $r$ is the radius of the circular path.
• By equating these forces, we get $qvB = \frac{mv^2}{r}$.
• Simplifying this equation gives $qB = \frac{mv}{r}$.
• The angular velocity ($\omega$) of the particle is related to its linear velocity and path radius by $\omega = \frac{v}{r}$.
• Substituting $\omega$ into the simplified force equation, we get $qB = m\omega$.
• The angular velocity is also related to the oscillator's linear frequency ($f$) by the equation $\omega = 2\pi f$.
• Substituting this relation, we find $qB = m(2\pi f)$.
• Rearranging the formula to solve for the magnetic field $B$, we get: $$ B = \frac{2\pi fm}{q} $$

It's important to note that the magnetic field strength required is independent of the radius of the dees. The radius determines the maximum energy the particle can achieve within the cyclotron.

Identifying Given Values

The problem provides the following information:

• Oscillator frequency, $f = 12 \times 10^6$ Hz.
• Dee radius, $R = 21$ inches. (Note: This value is not required for the calculation of magnetic field B).
• Particle type: Deuteron.
• Mass of deuteron, $m = 3.34 \times 10^{-27}$ kg.
• Charge of deuteron, $q = 1.6 \times 10^{-19}$ C.
• We will use the approximate value $\pi \approx 3.14159$.

Step-by-Step Calculation

We will now calculate the magnetic induction ($B$) using the formula derived above:

1. Formula: $$ B = \frac{2\pi fm}{q} $$
2. Substitute the given values: $$ B = \frac{2 \times \pi \times (12 \times 10^6 \text{ Hz}) \times (3.34 \times 10^{-27} \text{ kg})}{1.6 \times 10^{-19} \text{ C}} $$
3. Calculate the numerator part involving constants and frequency: $2 \times \pi \times 12 \times 10^6 \times 3.34 \times 10^{-27}$ $\approx 2 \times 3.14159 \times 12 \times 3.34 \times 10^{(6 - 27)}$ $\approx 251.441 \times 10^{-21}$
4. Perform the division: $$ B = \frac{251.441 \times 10^{-21}}{1.6 \times 10^{-19}} $$ $$ B = \frac{251.441}{1.6} \times 10^{(-21 - (-19))} $$ $$ B = 157.1506 \times 10^{-2} \text{ Tesla} $$
5. Express the final result: $$ B \approx 1.5715 \text{ Tesla} $$

Matching the Result to the Options

The calculated value for the magnetic field strength is approximately $1.5715$ Tesla. Let's compare this result to the provided options:

• 1. 1.573 tesla
• 2. 157.3 tesla
• 3. 15.73 tesla
• 4. 0.157 tesla

Our calculated value of $1.5715$ Tesla is very close to the value given in option 1 ($1.573$ Tesla). The slight difference can be attributed to the precision used for constants like $\pi$ or potentially minor variations in the provided mass or charge values.