Final answer:
The force on a 2 MeV proton moving perpendicular to a 2.5 tesla magnetic field is found by calculating the proton's velocity from its kinetic energy and then applying the magnetic force formula F = qvB, as the angle is 90 degrees making sin(theta) equal to 1.
Step-by-step explanation:
To find the force on a proton moving perpendicular to a magnetic field, we use the magnetic force equation F = qvBsin(\theta), where q is the charge of the proton, v is the velocity, B is the magnetic field strength, and \theta is the angle between the velocity and the magnetic field. In this case, the proton's charge is the elementary charge e = 1.6 \times 10^{-19} C, the velocity v can be found using the kinetic energy formula KE = \frac{1}{2}mv^2, where KE is the kinetic energy (2 MeV or 3.2 \times 10^{-13} joules) and m is the mass of the proton given as 1.65 \times 10^{-27} kg. Since the question states that the proton is moving perpendicular to the magnetic field, \theta is 90 degrees, making sin(\theta) equal to 1. Hence, the force is solely dependent on the charge, the velocity, and the magnetic field strength.
The steps to calculate this force are as follows:
- Convert the kinetic energy from MeV to joules if necessary.
- Calculate the velocity of the proton using the kinetic energy formula.
- Plug in the values into the magnetic force equation to find the force exerted on the proton by the magnetic field.
Applying the above steps, we find that the velocity v is \(\sqrt{\frac{2\times KE}{m}}\), and substituting v, q, and B into the equation, we get F = qvB. Thus, the force on a 2 MeV proton moving in a 2.5 T magnetic field is calculated using these values.