Respuesta :
Answer: Fmax = 5.54*10^-12 N
Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.
The maximum force experienced by a charge of magnitude (q) is given as
Fmax = qvB
Where v = velocity of electron.
The velocity of the electron can be gotten by using the work energy theorem.
The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.
mv²/2 = qV.
Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.
By substituting the parameters, we have that
(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000
(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2
v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31
v² = 64.36*10^(-16)/9.11×10^-31
v² = 7.0647×10^15
v = √7.0647×10^15
v = 8.40×10^7 m/s
Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41
Fmax = 5.54*10^-12 N
The magnitude of the maximum magnetic force that an electron can experience is Fmax = 5.54*10^-12 N.
Calculation of the magnitude:
Since the potential difference (V) =20kv = 20,000v
And strength of magnetic field (B) =0.41 T.
Now The maximum force should be
Fmax = qvB
here v = velocity of electron.
Now the kinetic energy be
mv²/2 = qV.
here
m = mass of an electronic charge = 9.11×10^-31 kg,
q = magnitude of an electronic charge = 1.609×10^-19 c,
v = velocity of electron,
V = potential difference = 20,000v.
So, the magnitude be
(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000
(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2
v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31
v² = 64.36*10^(-16)/9.11×10^-31
v² = 7.0647×10^15
v = √7.0647×10^15
v = 8.40×10^7 m/s
Now
Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41
Fmax = 5.54*10^-12 N
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