A cosmic-ray proton in interstellar space has an energy of 19.5 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 that region of space?

To find the magnetic field in the region of space where the cosmic-ray proton is executing a circular orbit, we can use the equation for the centripetal force acting on the proton. The centripetal force is provided by the magnetic force (q*v*B) acting on the proton, where q is the charge of the proton, v is its velocity, and B is the magnetic field.

The centripetal force is given by:

F = (mv^2)/r

Where m is the mass of the proton, v is its velocity, and r is the radius of the circular orbit.

Since the proton is in circular motion, its velocity can be calculated using the equation:

v = ω*r

Where ω is the angular velocity of the proton.

In circular motion, the angular velocity is related to the energy and mass of the particle by the equation:

E = (1/2) * m * ω^2 * r^2

Rearranging the equation, we can solve for ω:

ω = sqrt(2 * E / (m * r^2))

Substituting the value of ω into the equation for velocity, we have:

v = sqrt(2 * E / (m * r))

Now we can substitute the value of v into the equation for the centripetal force:

F = (m * (sqrt(2 * E / (m * r))))^2 / r

Simplifying, we have:

F = (2 * E) / r

The magnetic force acting on the proton is given by:

F = q * v * B

Setting the equations for the centripetal force and the magnetic force equal to each other, we can solve for the magnetic field B:

(2 * E) / r = q * v * B

B = (2 * E) / (q * v * r)

Now we can substitute the given values into the equation to calculate the magnetic field:

E = 19.5 MeV = 19.5 * 10^6 eV
q = charge of a proton = 1.602 x 10^-19 C
v = velocity of the proton = sqrt(2 * E / m)
r = radius of the orbit = 2.28 * 10^11 m

First, we need to find the mass of the proton using Einstein's equation E = mc^2:

E = mc^2
m = E / c^2
m = (19.5 * 10^6 eV) / (c^2)

The speed of light, c, is approximately 3 x 10^8 m/s. So:

m = (19.5 * 10^6 eV) / ((3 x 10^8 m/s)^2)

Now we can substitute the value of m into the equation for the velocity of the proton:

v = sqrt(2 * E / m)

Finally, we can substitute all the values into the equation for the magnetic field to find the answer.

To determine the magnetic field in the region of space, we can use the equation for the radius of a charged particle in a magnetic field. The equation is given by:

r = (m*v)/(e*B),

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

In this case, we can assume that the charged particle is a proton, which has a charge of e = 1.6 x 10^-19 C and a mass of m = 1.67 x 10^-27 kg. We are also given the radius, r = 2.28 × 10^11 m, and the energy of the proton, 19.5 MeV.

To find the velocity of the proton, we can use the relativistic energy-momentum relation:

E^2 = (mc^2)^2 + (pc)^2,

where E is the total energy, m is the mass of the particle, c is the speed of light, and p is the momentum.

First, we need to convert the proton's energy from MeV to joules:

19.5 MeV * (1.6 x 10^-13 J/MeV) = 3.12 x 10^-12 J.

Using the equation for the total energy, we can solve for the magnitude of the momentum:

E^2 = (mc^2)^2 + (pc)^2,
(3.12 x 10^-12 J)^2 = (1.67 x 10^-27 kg * (3.0 x 10^8 m/s)^2) + (p)^2,
(3.12 x 10^-12 J)^2 - (1.67 x 10^-27 kg * (3.0 x 10^8 m/s)^2) = (p)^2,
p = (3.12 x 10^-12 J)^2 - (1.67 x 10^-27 kg * (3.0 x 10^8 m/s)^2)^0.5,
p ≈ 2.87 x 10^-18 kg*m/s.

The momentum can also be written as p = m*v, where v is the velocity of the proton. Therefore,

v = p / m = (2.87 x 10^-18 kg*m/s) / (1.67 x 10^-27 kg) ≈ 1.72 x 10^9 m/s.

Now we have all the values needed to calculate the magnetic field:

r = (m * v) / (e * B),
2.28 x 10^11 m = ((1.67 x 10^-27 kg) * (1.72 x 10^9 m/s)) / (1.6 x 10^-19 C) * B,
2.28 x 10^11 m * (1.6 x 10^-19 C) = (1.67 x 10^-27 kg) * (1.72 x 10^9 m/s) * B,
B = (2.28 x 10^11 m * (1.6 x 10^-19 C)) / ((1.67 x 10^-27 kg) * (1.72 x 10^9 m/s)).

Evaluating the above expression:

B ≈ 1.31 x 10^-10 Tesla.

Therefore, the magnetic field in that region of space is approximately 1.31 x 10^-10 Tesla.