A. using force balance
Fe = Fm
qE = qVB
E = VB
V = E/B = 187500/0.125
V = 1.5*10^6 m
Now using circular motion
Fc = Fm
mV^2/r = qVB
q/m = V/rB
r = d/2 = 0.2505/2
q/m = 1.5*10^6*2/(0.2505*0.125)
q/m = 9.58*10^7 C/kg
B.
for proton
q/m = 1.6*10^-19/(1.67*10^-27) =9.58*10^7
for electron
q/m = 1.6*10^-19/(9.1*10^-31) = 1.758*10^11
for neutron
q/m = 0/(1.67*10^-24) = 0
So correct answer will be proton.
An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500...
An unknown particle moves in a straight line through crossed electric and magnetic fields with E = 1.5 kV/m and B = 0.034 T. If the electric field is turned off, the particle moves in a circular path of radius r = 2.7 cm.
8. [8 pts] Crossed electric and magnetic fields are established over a certain region. Derive an equation, in terms of the electric and magnetic field strength, for the speed of an electron traveling through the crossed-fields without deflection.
Crossed electric and magnetic fields are established over a certain region. The magnetic field is 0.6000 T vertically downward. The electric field is 2.000 x 106 V/m horizontally east. An electron, traveling horizontally northward, experiences zero resultant force from these fields and so continues in a straight line. What is the electron's speed? m/s
1. A stream of electrons passes through a velocity filter where the crossed magnetic and electric fields are O.02 T and 5.0 x10 V/m respectively. Find the kinetic energy (in electron volts ) of the electrons passing through the filter. ( 1 eV 1.60 X 10-19 J).
elp with physics 102 An electron moves through a region of crossed electric and magnetic fields. The electric field E = 2 000 V/m and is directed straight down. The magnetic field B = 0.80 T and is directed to the left. Specify the directions of electric and magnetic forces on the electron if it moves into the paper. For what velocity v of the electron will the electric force exactly cancel the magnetic force? A rod(length = 10 cm)...
Answer should be V = (E)/ (B) 6 Suppose a charged particle is moving through a region of space in which there is an electric field perpendicular to its velocity vector, and also a magnetic field perpendicular to both the particle's velocity vector and the electric field. Show that there will be one particular velocity at which the particle can be moving that results in a total force of zero on it; this requires that you analyze both the magnitudes...
A charged particle is moving in a uniform, constant magnetic field. Which one of the following statements concerning the magnetic force exerted on the particle is false It does no work on the particle. It increases the speed of the particle. It changes the velocity of the particle. It can act only on a particle in motion. It does not change the kinetic energy of the particle. A circular current loop with radius of 0.100 m is located in the...
Electric Fields Equipment and Setup: Mathematica file- ElectricFields.nb Section A: Electric Fields Due to Two Charges Computer Setup for Section A 1. The first interactive panel shows electric fields due to two point charges, Qat (-1 m,0) and Q, at (1 m,0). The controls for this panel are at the top on the left 2. The top line has two checkboxes: one to Show Axes and the other to Show Field Lines. The top line also has a slider labeled...
Pls help with all questions a) A charged particle travels undeflected through perpendicular electric and magnetic fields whose magnitudes are 4000 N/C and 20mT, respectively. Find the speed of the particle if it is an alpha particle. (Alpha particle is a helium nucleus-a positive ion with a double positive charge of +2e.) I don't understand how the answer is v= 2.0*10^5 m/s. How could the velocity be independent of the charge since v = F / qBsinθ and q =2e?...