Magnitude and direction of magnetic field Feyman is performing an experiment whe

Question

Magnitude and direction of magnetic field Feyman is performing an experiment where he accelerates an electron from rest over a distance of 10.0 cm through a region of constant electric field with magnitude of 2.50 Times 1064 V/m. The electron then enters a region where there is a uniform 2.0 T magnetic field. In which direction, with respect to the electric field, is the electron moving? What are the maximum and minimum magnitudes of the magnetic force acting on this electron? If the magnitude of the magnetic force is 3.00 Times 10^-12 N, what is the smallest angle it entered the magnetic field with?

Explanation / Answer

Here,

Mass of electron = 9.10 * 10^-31 Kg
Charge of Electron = 1.60 * 10^-19 C

Electric Field = 2.50*10^4 V/m
Magnatic Field = 2 T

Distance = 10cm = .1m

part A:
as charge on electron is positive so it will be being in directin of Electric field

Part B:
from lortez law
F = qvB sinA

where F = Q*E = 0.5*MV^2

1.60*10^-19 * 2.50*10^4 =0.5*9.10 * 10^-31*V^2

V = 9.37 * 10^7 m/s

Therefore,
for max Force SinA = 1 or A = 90 degrees
Fmax = 1.60*10^-19 * 9.37 * 10^7 * 2 * sin90
Fmax = 2.99 * 10^-11 N

for min Force SinA = 0 or A = 0 degrees
Fmin = 1.60*10^-19 * 2.09 * 10^7 * 2 * sin0
Fmin = 0 N

The vaue of Max and Min force of eletron while moving is 2.99 * 10^-11 N and 0N

part C:
F = 3*10^-12 N
as
F = Q*V*B*SinA

SinA = F /( Q*V*B)
SinA = 3*10^-12 / ( 1.60*10^-19 * 9.37 * 10^7 * 2 )

Sin A = 0.100
A = arcSin(0.100)
A = 5.73 degrees

The minimum angle for electron when it entered in magnatic Field is 5.73 degrees

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