Question

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?

Answer 1

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

Step-by-step 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

`v = \sqrt{(2K.E)/(m) }`

where;

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

`v = \sqrt{(2*3.12*10^(-12))/(1.67*10^(-27)) } \\\\v = 6.1127*10^7 \ m/s`

Step 2:

determine the magnetic field strength in that region of space

magnetic force = centripetal force

`qvB = (mv^2)/(r) \\\\B = (mv)/(rq)\\\\`

Where;

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

`B = (mv)/(rq) = ((1.67*10^(-27))(6.1127*10^7))/((2.28*10^(11))(1.6*10^(-19)))\\\\B =2.7983 *10^(-12) \ T`

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

Answer 2

B = (2.80 × 10⁻¹²) T

Step-by-step 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!!!