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
Answer:
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.
