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?

Question

Itzel Wyatt

Physics

2 March, 07:42

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?

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Answers (2)

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Guillermo Osborne · Answer 1

Answer: 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

Mattie Hodge · Answer 2

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.