A proton is accelerated through a potential difference of 250V. It then enters a uniform magnetic field and moves in a circular path of radius 10.0cm. What is the magnitude of the magnetic field? What is the period of revolution for the circling proton?

a) 7.52 ×10^-4 T, b) 4.72×10^-7s

Explanation: by using the work energy thereom, the work done on the electron by the voltage source (potential) equals the kinetic energy.

qV =mv^2/2

Where q = magnitude of electronic charge = 1.609×10^-19 c

V = potential difference = 250v

m = mass of an electronic charge = 9.11×10^-31 kg

v = velocity of electron =?

By substituting the parameters, we have that

1.609×10^-19 × 250 = 9.11×10^-31 × v^2/2

1.609×10^-19 × 250 ×2 = 9.11×10^-31 ×v^2

v^2 = 1.609×10^-19 × 250 ×2/ 9.11×10^-31

v^2 = 8.05×10^-17/ 9.11×10^-31

v^2 = 1.77×10^14

v = √1.77×10^14

v = 1.33×10^7 m/s

The centripetal force required for the motion of the electron is gotten from the magnetic force on the electron.

qvB = mv^2/r

By cancelling "v" on both sides of the equation, we have that

qB = mv/r

Where r = radius = 10cm = 0.1m

1.609×10^-19 × B = 9.11 ×10^-31 ×1.33×10^7/ 0.1

B = (9.11 ×10^-31 × ×1.33×10^7)/ (0.1 ×1.609×10^-19)

B = 1.21×10^-23/ 1.609×10^-20

B = 0.752×10^-3

B = 7.52 ×10^-4 T

To get the period of motion, we recall that

v = ωr

Where ω = angular frequency

1.33×10^7 = ω×0.1

ω = 1.33×10^7/ 0.1

ω = 13.3×10^7

ω = 1.33×10^6 rad/s

But the period (T) of a periodic motion is defined as

T = 2π/ω

T = 2 × 3.142 / 1.33×10^6

T = 4.72×10^-7s

B = 0.0228T

T= 2.87×10-⁶s

Explanation:

Given V = 250V, R = 10cm = 0.1m

q = 1.6×10-¹⁹C, m = 1.67×10-²⁷kg

From the work energy theorem the electric energy of the field = kinetic energy of the proton.

qV = 1/2×mv²

v = √(2qV/m) = √(2×1.6×10-¹⁹×250/(1.67×10-²⁷))

v = 218870m/s

The magnetic force on the proton by newton's 2nd law of motion is related as

qvB = ma = m× v²/R

Rearranging

B = mv/qR = 1.67×10-²⁷ × 218870/(1.6×10-¹⁹ × 0.1) = 0.0228T

T = 2π/ω = 2πR/v = 2π×0.1/218870 = 2.87×10-⁶ = 2.87μs.