A cosmic-ray proton in interstellar space has an energy of 12.5 MeV and executes a circular orbit having a radius equal to that of Mars' orbit around the Sun (2.28 ✕ 1011 m). What is the magnetic field (in T) in that region of space?

Answer 1

We can use the formula for the radius of curvature of a charged particle's trajectory in a magnetic field:

r = (mv) / (qB),

where r is the radius of the trajectory, m is the mass of the particle, v is its velocity, q is its charge, and B is the magnetic field strength.

First, we need to find the mass and the velocity of the proton. The mass of a proton is approximately:

m = 1.67 * 10^-27 kg

The energy of the proton is given as 12.5 MeV, which is equivalent to:

E = 12.5 * 10^6 * 1.6 * 10^-19 J = 2.0 * 10^-12 J

For a non-relativistic proton, we can relate its kinetic energy (E) to its mass (m) and velocity (v) as follows:

E = (1/2) * m * v^2

Now, we can solve for v:

v^2 = (2 * E) / m
v = sqrt((2 * 2.0 * 10^-12) / (1.67 * 10^-27))
v ≈ 7.66 * 10^7 m/s

Now, we can find the magnetic field strength (B) using the formula for the radius of curvature:

B = (mv) / (qr)

The charge of a proton (q) is approximately:

q = 1.6 * 10^-19 C

Now, we can calculate the magnetic field strength:

B = (1.67 * 10^-27 * 7.66 * 10^7) / (1.6 * 10^-19 * 2.28 * 10^11)
B ≈ 3.19 * 10^-10 T

Therefore, the magnetic field in that region of space is approximately 3.19 * 10^-10 T.