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?

B = (2.80 × 10⁻¹²) T

Explanation:

First of, we convert the 19.5 MeV to Joules

19.5 MeV = 19.5 × 10⁶ × 1.602 × 10⁻¹⁹

= (3.124 × 10⁻¹²) J

The velocity of the cosmic-ray proton can be calculated from the kinetic energy formula

E = (1/2) mv²

m = mass of a proton = (1.67 × 10⁻²⁷) kg

(3.124 × 10⁻¹²) = (1/2)(1.67 × 10⁻²⁷)v²

v = (6.117 × 10⁷) m/s

And since the magnetic force keeps the cosmic ray proton in uniform circular motion,

Magnetic force = Centripetal force keeping the proton in circular motion.

qvB = (mv²/r)

q = charge on a proton = (1.602 × 10⁻¹⁹) C

v = velocity of the proton = (6.117 × 10⁷) m/s

B = magnetic field = ?

r = radius of circular motion = (2.28 × 10¹¹) m

B = (mv/qr)

B = (1.67 × 10⁻²⁷ × 6.117 × 10⁷) ÷ (1.602 × 10⁻¹⁹ × 2.28 × 10¹¹)

B = (2.797 × 10⁻¹²) T

Hope this Helps!!!

The magnetic field strength in that region of space is 2.7983 x 10⁻¹² T

Explanation:

Given;

Energy of the cosmic-ray proton, E = 19.5 MeV = 19.5 x 10⁶ x 1.6 x 10⁻¹⁹

                                                       E = 3.12 x 10⁻¹² J

Radius of the circular orbit, r = 2.28 x 10¹¹ m

Step 1:

determine the speed of cosmic-ray proton

Kinetic energy of the cosmic-ray proton;

K.E = ¹/₂mv²

where;

m is mass of proton, m = 1.67 x 10⁻²⁷ kg

Step 2:

determine the magnetic field strength in that region of space

magnetic force = centripetal force

Where;

q is charge of proton, q = 1.6 x 10⁻¹⁹ C

Therefore, the magnetic field strength in that region of space is 2.7983 x 10⁻¹² T

it would be c. predictability