Let's determine the kinetic energy step by step.

The maximum speed a proton reaches in a cyclotron (when it has just reached the outer edge of the dee with radius $$r$$) is given by:

$$v = \frac{qBr}{m}$$

where

$$q = 1.60 \times 10^{-19}$$ C (charge of a proton),

$$B = 0.66$$ T (magnetic field),

$$r = 0.60$$ m (dee radius),

$$m = 1.67 \times 10^{-27}$$ kg (mass of a proton).

The kinetic energy (non-relativistic) is:

$$K = \frac{1}{2} m v^2$$

Substituting the expression for $$v$$:

$$K = \frac{1}{2} m \left(\frac{qBr}{m}\right)^2 = \frac{q^2B^2r^2}{2m}$$

Plug in the numbers:

Calculate $$q^2$$:

$$q^2 = (1.60 \times 10^{-19})^2 = 2.56 \times 10^{-38}\ \text{C}^2$$

Calculate $$B^2$$:

$$B^2 = (0.66)^2 \approx 0.4356\ \text{T}^2$$

Calculate $$r^2$$:

$$r^2 = (0.60)^2 = 0.36\ \text{m}^2$$

Now, substitute into the kinetic energy formula:

$$K = \frac{2.56 \times 10^{-38} \times 0.4356 \times 0.36}{2 \times 1.67 \times 10^{-27}}$$

First, compute the numerator:

Multiply the numbers:

$$0.4356 \times 0.36 \approx 0.1568$$

$$2.56 \times 10^{-38} \times 0.1568 \approx 4.01 \times 10^{-39}$$

Then, compute the denominator:

$$2 \times 1.67 \times 10^{-27} = 3.34 \times 10^{-27}$$

So,

$$K \approx \frac{4.01 \times 10^{-39}}{3.34 \times 10^{-27}} \approx 1.20 \times 10^{-12}\ \text{J}$$

Convert the kinetic energy from joules to electron volts (eV). Recall that:

$$1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$$

Therefore,

$$K \approx \frac{1.20 \times 10^{-12}\ \text{J}}{1.60 \times 10^{-19}\ \text{J/eV}} \approx 7.5 \times 10^6\ \text{eV} = 7.5\ \text{MeV}$$

Comparing with the options provided, the closest answer is:

Option C: 7 MeV

Thus, the kinetic energy of the proton beam produced by the cyclotron is approximately 7 MeV.

Moving Charges And Magnetism

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