Question

A 2MeV proton is moving perpendicular to a uniform magnetic field of 2.5 T.the force on a proton is

Answer 1

8*10^-12

Step-by-step 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

Answer 2

7.8x10-12N

Step-by-step 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