Proton \(A\) is moving with speed \(10^6 \mathrm{~m} / \mathrm{s}\) in a magnetic field of \(0.01 \mathrm{~T}\). Proton \(B\) is moving in the same magnetic field with speed \(2 \times 10^6 \mathrm{~m} / \mathrm{s}\). One can conclude that (A) Both protons have an orbital radius of about 1 meter. (B) Both protons have the same orbital period of about \(6.5 \times 10^{-6} \mathrm{~s}\). (C) proton \(B\) has twice the orbital period of proton \(A\). (D) proton \(B\) has twice the orbital frequency of proton \(A\). (E) proton \(B\) has half the orbital radius of proton \(B\).

In a magnetic field, the force acting on a moving charged particle can be determined by the Lorentz force formula: \(F = q(vB\sin{\theta})\). For a charged particle moving in a circular orbit within a uniform magnetic field, the magnetic force provides the centripetal force. The centripetal force is given by \(F_{c} = \frac{mv^2}{r}\), where \(m\) is mass, \(v\) is speed, and \(r\) is the orbit radius of the particle. Therefore, we equate the forces: \(q(vB\sin{\theta}) = \frac{mv^2}{r}\). Since both protons are moving in the same magnetic field, \(\theta = 90^{\circ}\), so \(\sin{\theta} = 1\). Now, we want to find a formula for radius \(r\).

To find the formula for \(r\), rearrange the equation from step 1: \(r = \frac{mv}{qB}\). It's important to note that the charge of a proton is \(q = 1.6 \times 10^{-19}\ \mathrm{C}\) and the mass of a proton is \(m = 1.67 \times 10^{-27}\ \mathrm{kg}\). Plug these values along with the speed and magnetic field of protons A and B, and calculate the radius for each proton: For proton A: \(r_A = \frac{(1.67 \times 10^{-27}\ \mathrm{kg})(10^6\ \mathrm{m/s})}{(1.6 \times 10^{-19}\ \mathrm{C})(0.01\ \mathrm{T})} \approx 1.04\ \mathrm{m}\) For proton B: \(r_B = \frac{(1.67 \times 10^{-27}\ \mathrm{kg})(2 \times 10^6\ \mathrm{m/s})}{(1.6 \times 10^{-19}\ \mathrm{C})(0.01\ \mathrm{T})} \approx 2.08\ \mathrm{m}\)

Now we will calculate the orbital period \(T\) and the frequency \(f\) of each proton. Note that the orbital period is the time for one orbit and is given by \(T = \frac{2\pi r}{v}\), while the frequency is the number of orbits per second, which can be calculated by \(f = \frac{1}{T}\). Calculate these values for protons A and B: For proton A: \(T_A = \frac{2\pi (1.04\ \mathrm{m})}{10^6\ \mathrm{m/s}} \approx 6.57 \times 10^{-6}\ \mathrm{s}\) and \(f_A = \frac{1}{6.57 \times 10^{-6}\ \mathrm{s}} \approx 152056\ \mathrm{Hz}\) For proton B: \(T_B = \frac{2\pi (2.08\ \mathrm{m})}{(2 \times 10^6\ \mathrm{m/s})} \approx 6.57 \times 10^{-6}\ \mathrm{s}\) and \(f_B = \frac{1}{6.57 \times 10^{-6}\ \mathrm{s}} \approx 152056\ \mathrm{Hz}\)

Reviewing the calculations from steps 2 and 3, we can compare the results of both protons A and B to identify the correct option. We found that: 1. Both protons have different orbital radii. 2. Protons A and B have the same orbital period of about \(6.57 \times 10^{-6}\ \mathrm{s}\). 3. Protons A and B have the same orbital frequency of about 152056 Hz. Therefore, the correct answer is: (B) Both protons have the same orbital period of about \(6.5 \times 10^{-6}\ \mathrm{s}\).