Question

Protons in the solar wind reach Earth's magnetic field with a speed of \(400 \mathrm{~km} / \mathrm{s}\). If the magnitude of this field is \(5.0 \cdot 10^{-5} \mathrm{~T}\) and the velocity of the protons is perpendicular to it, what is the cyclotron frequency of the protons after entering the field? a) \(122 \mathrm{~Hz}\) b) \(233 \mathrm{~Hz}\) c) \(321 \mathrm{~Hz}\) d) \(432 \mathrm{~Hz}\) e) \(763 \mathrm{~Hz}\)

Short answer

Answer: The cyclotron frequency of the protons is approximately 321 Hz.

Step-by-step solution

Convert velocity to m/s

We're given the velocity of the protons as \(400\,km/s\). To convert this to m/s, multiply by 1000: \(v = 400\,km/s \times \frac{1000\,m}{1\,km} = 400000\,m/s\)

Calculate the cyclotron frequency

Using the given values and the formula for cyclotron frequency, we can find the frequency as follows: \(f_c = \frac{qB}{2\pi m} = \frac{(1.6 \times 10^{-19}\,C)(5.0 \times 10^{-5}\,T)}{2\pi (1.67 \times 10^{-27}\,kg)}\) Now we just plug in the values and solve for \(f_c\): \(f_c = \frac{(1.6 \times 10^{-19}\,C)(5.0 \times 10^{-5}\,T)}{2\pi (1.67 \times 10^{-27}\,kg)} \approx 321\,Hz\) So the cyclotron frequency of the protons is approximately \(321\,Hz\), which corresponds to answer choice (c).