Answer:
= 7.88 × 10^-12 T
Explanation:
From the above question, we are told that:
Kinetic Energy of the proton is K. E = 10.0 MeV
Step 1
We convert 10.0 MeV to Joules
1 Mev = 1.602 × 10-13 Joules
10.0 MeV = 10.0 × 1.602 × 10^-13 Joules = 1.602 × 10^-12 J
Hence, the Kinectic energy of a proton = 1.602 × 10^-12 J
Step 2
Find the Speed of the Proton
The formula for Kinectic Energy =
K.E = 1/ 2 mv²
Where
m = mass of the proton
v = speed of the proton
K.E of the proton = 1.602 × 10^-12 J
Mass of the proton = 1.6726219 × 10^-27 kilograms
Speed of the proton = ?
1.602 × 10^-12J = 1/2 × 1.6726219 × 10^-27 × v²
1.602 × 10^-12J = 8.3631095 ×10^-28 × v²
v² = 1.602 × 10^-12/8.3631095 ×10^-28
v = √(1.602 × 10^-12/8.3631095 ×10^-28)
v = 43772331.227m/s
v = 4.3772331227 × 10^7m/s
Approximately = 4.4 × 10^7 m/s
Step 3
Find the Magnetic Field of that region of space
The formula for Magnetic Field =
B = m v / q r
We are told that the proton executes a circular orbit, hence,
mv = √2m(KE)
m = Mass of the proton = 1.6726219 × 10^-27 kg
K.E of the proton = 1.602 × 10^-12 J
v = speed of the proton = 4.4 × 10^7 m/s
q = Electric charge = 1.6 × 10^-19 C
r = radius of the orbit = 5.80Ã10^10 m
= 5.8 × 10^10m
Magnetic Field =
=√ (2 × 1.6726219 × 10^-27 kg × 1.602 × 10^-12 J) /( 1.6 × 10^-19 C × 5.80 × 10^10 m)
= 7.88 × 10^-12 T
The magnetic field in that region of space is approximately 7.88 × 10^-12 T
