Question

A uniform magnetic ficld points north; its magnitude is 1.5 T. A proton with kinetic energy \(8.0 \times 10^{-13} \mathrm{J}\) is moving vertically downward in this field. What is the magnetic force acting on it?

Short answer

Answer: The magnetic force acting on the proton is approximately \(2.35 \times 10^{-11} \mathrm{N}\).

Step-by-step solution

Find the velocity of the proton

First, we need to find the velocity of the proton using its given kinetic energy. The kinetic energy of the proton can be calculated using the following formula: \(K = \frac{1}{2}mv^2\) Where \(K\) is the kinetic energy, \(m\) is the mass of the proton, and \(v\) is the velocity of the proton. We know that the kinetic energy of the proton is \(8.0 \times 10^{-13} \mathrm{J}\), and the mass of the proton is approximately \(1.67 \times 10^{-27} \mathrm{kg}\). We can now solve for the velocity, \(v\): \(8.0 \times 10^{-13} \mathrm{J} = \frac{1}{2}(1.67 \times 10^{-27} \mathrm{kg})v^2\)

Solve for the proton's velocity

To find the velocity of the proton, we will need to rearrange the equation and solve for \(v\): \(v^2 = \frac{2(8.0 \times 10^{-13} \mathrm{J})}{1.67 \times 10^{-27} \mathrm{kg}}\) \(v^2 \approx 9.58 \times 10^{14} \mathrm{m^2/s^2}\) \(v \approx 9.79 \times 10^7 \mathrm{m/s}\) Now that we have the velocity of the proton, we can proceed to find the magnetic force acting on it.

Calculate the magnetic force on the proton

We can find the magnetic force acting on the proton using the following formula: \(F = qvB\sin{\theta}\) where \(F\) is the magnetic force, \(q\) is the charge of the proton, \(v\) is the proton's velocity, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the velocity vector and the magnetic field vector. The charge of a proton is \(q = 1.6 \times 10^{-19} \mathrm{C}\), the magnetic field strength is \(B = 1.5 \mathrm{T}\), and the proton's velocity is approximately \(9.79 \times 10^7 \mathrm{m/s}\). Since the proton is moving vertically downward and the magnetic field points north, the angle between the velocity vector and the magnetic field vector is \(\theta = 90^{\circ}\) or \(\frac{\pi}{2}\) radians. Therefore: \(F = (1.6 \times 10^{-19} \mathrm{C})(9.79 \times 10^7 \mathrm{m/s})(1.5 \mathrm{T})\sin{\frac{\pi}{2}}\)

Calculate the magnetic force

Finally, we can compute the magnetic force acting on the proton in the uniform magnetic field: \(F \approx 2.35 \times 10^{-11} \mathrm{N}\) Hence, the magnetic force acting on the proton is approximately \(2.35 \times 10^{-11} \mathrm{N}\).