Answer:
7.8x10-12N
Explanation:
We know that
Magnetic force = F = qVB
And
Also Kinetic energy K.E is
E = (1/2)mV²
So making v subject
V = √(2E / m)
And
E = KE = 2MeV
= 2 × 106 eV
= 2 × 106 × 1.6 × 10–19 J
= 3.2 × 10–13 J
And then
V= √2x3.2E-13/1.6E-27
1.9E7m/s
Given that
mass of proton = 1.6 × 10–27 kg,
Magnetic field strength B = 2.5 T.
So F= qBv sinစ
=
So F = 1.6 × 10–19 × 2.5 × 1.9 x10^7 x sin 90°
= 7.8 x 10^-12N
Answer:
8*10^-12
Explanation:
Given that
Energy of proton, K = 2 MeV = 2 * 1.6*10^-19 *10^6 = 3.2*10^-13
magnetic field strength, B = 2.5 T
mass of proton, m = 1.67*10^-27 kg
K = ½mv², making v² the subject of formula by rearranging, we have
v² = 2k/m
v² = (2 * 3.2*10^-13) / 1.67*10^-27
v² = 6.4*10^-13 / 1.6*10^-27
v² = 4*10^14
v = √4*10^14
v = 2*10^7 m/s
f = qvbsinθ, where
θ = 90
v = 2*10^7 m/s
b = 2.5 T
q = 1.6*10^-19
f = 1.6*10^-19 * 2*10^7 * 2.5 sin 90
f = 8*10^-12 N
thus, the force on the proton is 8*10^-12
