Super magnetic fields and astronomical units!?

Question

An ionized hydrogen atom (that is, a free proton) is moving through the magnetic field of the Milky Way galaxy, at a location distant from any star. The kinetic energy of the proton is 5.60 keV, and it moves through a circular path having a radius of 0.0317 astronomical units. An astronomical unit (AU) equals the average radius of the Earth's orbit around the Sun, 1.50×1011 m. What is the strength of the galactic magnetic field in this region of space?

Answer 1

The kinetic energy of the proton (in Joules) is:

K = 5,600 eV x (1.60 x 10^-19 J) = 8.96 x 10^-16 J

The velocity of the proton is:

v = (2K / m)^(1/2) = ((2 x (8.96 x 10^-16 J)) / (1.67 x 10^-27 kg))^(1/2) = 1.04 x 10^6 m/s

The radius of the circular path is:

r = 0.0317 x (1.50 x 10^11 m) = 4.76 x 10^9 m

The magnetic field magnitude is:

B = mv / qr = ((1.67 x 10^-27 kg) x (1.04 x 10^6 m/s)) / ((1.60 x 10^-19 C) x (4.76 x 10^9 m))

= 2.28 x 10^-12 T

Answer 2

my bf graduated in physics he says ... zero

yeah sorry about that, i thought he may have been lying.