A cosmic-ray proton in interstellar space has an energy of 59 MeV and executes a circular orbit with a radius equal to that of Mercury’s orbit (5.8 × 1010 m) around the Sun. What is the magnetic field in that region of space? The proton has a charge of 1.60218 × 10−19 C and a mass of 1.67262 × 10−27 kg. Answer in units of T

Explanation:

Let mass and velocity of proton be m and v .

1/2 m v² = 59 x 10⁶ e V

= 59 x 10⁶ x 1.6 x 10⁻¹⁹ J

= 94.4 x 10⁻¹³ J

mv² = 188.8 x 10⁻¹³ J

v² =  188.8 x 10⁻¹³ / m

= 188.8 x 10⁻¹³ / 1.67262 x 10⁻²⁷

= 112.8768 x 10¹⁴

v = 10.62 x 10⁷ m / s

In circular path of proton , magnetic force equals centripetal force .

m v² / r = B q v , B is magnetic field , q is charge on proton , r is radius of circular path .

188.8 x 10⁻¹³ / 5.8 x 10¹⁰ = B x 1.6 x 10⁻¹⁹ x  10.62 x 10⁷

B = 1.9157 x 10⁻¹¹ T.

answer: (431,2)

step-by-step explanation:

1/0.14nm = 10^10/1.4 a lot