Respuesta :
Answer:
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!!!
Answer:
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
[tex]v = \sqrt{\frac{2K.E}{m} }[/tex]
where;
m is mass of proton, m = 1.67 x 10⁻²⁷ kg
[tex]v = \sqrt{\frac{2*3.12*10^{-12}}{1.67*10^{-27}} } \\\\v = 6.1127*10^7 \ m/s[/tex]
Step 2:
determine the magnetic field strength in that region of space
magnetic force = centripetal force
[tex]qvB = \frac{mv^2}{r} \\\\B = \frac{mv}{rq}\\\\[/tex]
Where;
q is charge of proton, q = 1.6 x 10⁻¹⁹ C
[tex]B = \frac{mv}{rq} = \frac{(1.67*10^{-27})(6.1127*10^7)}{(2.28*10^{11})(1.6*10^{-19})}\\\\B =2.7983 *10^{-12} \ T[/tex]
Therefore, the magnetic field strength in that region of space is 2.7983 x 10⁻¹² T
