Question

A 2 Mev proton is moving perpendicular to a uniform magnetic field of \(2.5\) tesla. The force on the proton is (a) \(3 \times 10^{-10} \mathrm{~N}\) (b) \(70.8 \times 10^{-11} \mathrm{~N}\) (c) \(3 \times 10^{-11} \mathrm{~N}\) (d) \(7.68 \times 10^{-12} \mathrm{~N}\)

Short answer

The force on the proton is approximately \(4.34 \times 10^{-12}\,\text{N}\). However, none of the given options match this result, which may indicate an error in the options provided.

Step-by-step solution

Convert energy to Joules

Given the energy of the proton is 2 MeV (Mega electron volts), we will convert it to Joules using the following conversion factor: 1 eV = \(1.6 \times 10^{-19}\) J 1 MeV = \(10^6\) eV So, \(2\,\text{MeV} = 2 \times 10^6 \times 1.6 \times 10^{-19}\, \text{J} = 3.2 \times 10^{-13}\,\text{J}\)

Calculate the velocity of the proton

To calculate the velocity of the proton, we need to consider the kinetic energy formula, \(E = \frac{1}{2}mv^2\), where \(E\) is the kinetic energy, \(m\) is the mass of the proton, and \(v\) is the velocity of the proton. The mass of a proton, \(m = 1.67 \times 10^{-27}\,\text{kg}\). Rearrange the formula to find the velocity: \(v = \sqrt{\frac{2E}{m}}\) Substitute the values: \(v = \sqrt{\frac{2 \times 3.2 \times 10^{-13}\text{J}}{1.67 \times 10^{-27}\,\text{kg}}}\) Calculating the velocity, we get: \(v = 1.082 \times 10^{7}\, \text{m/s}\)

Calculate the force on the proton

Now, we can calculate the force on the proton using the Lorentz force formula: \(F = qvB\). The charge of a proton, \(q = 1.6 \times 10^{-19}\,\text{C}\), and the magnetic field strength, \(B = 2.5\,\text{T}\). Substitute the values: \(F = (1.6 \times 10^{-19}\text{C}) \times (1.082 \times 10^7\,\text{m/s}) \times (2.5\, \text{T})\) Calculating the force, we get: \(F \approx 4.34 \times 10^{-12}\,\text{N}\) Comparing our answer to the given options, it seems none of them match our result, which might indicate that there might be a mistake in the options given in the exercise.