Question

q1

A charged particle is injected at 213 m/s into a 0.0649-T uniform magnetic magnetic field perpendicularly to the field. The diameter of its orbit is measured and found to be 0.0405 m. What is the charge-to-mass ratio of this particle?

Answer 1

diameter d = 2*r = 2*(m*v)/(q*B).....

q/m = (2*v)/(B*d) = 2*213/(0.0649*0.0405) = 162072.704446 C/kg = 1.62*10^5 C/kg

Answer 2

radius of the orbit

r=d/2=0.0405/2=0.02025 m

centripetal force =magnetic force

qvB=mv²/r

q/m =v/rB =213/0.02025*0.0649

q/m=1.62*10⁵ C/kg or 162072.7 C/kg

Answer 3

e/m = v / B*r = 213 / (0.0649*0.0405) C/kg = 8.1*10^4 C/kg

Answer 4

Circular Path from Magnetic Field

If a charge moves into a magnetic field with direction perpendicular to the field, it will follow a circular path. The magnetic force, being perpendicular to the velocity, provides the centripetal force.

Radius of path produced by magnetic field 2 mv r mv 2 If the velocity v is produced by an accelerating voltage V Substitution gives Magnetic field 1 2mV out toward observer

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Q - A charged particle is injected at 213 m/s into a 0.0649-T uniform magnetic magnetic field perpendicularly to the field. The diameter of its orbit is measured and found to be 0.0405 m. What is the charge-to-mass ratio of this particle?

Answer : here

r=0.0405/2 m = 0.02025

B=0.0649T

v=213m/s

Thus , putting them in the equation, we get :

$r=\frac{mv}{qB}$

$=>0.02025=\frac{m\times 213}{q\times 0.0649}$

$=>\frac{q}{m}=\frac{213}{0.02025\times 0.0649}=162072.7=1.62\times 10^5 C/kg$