Singly ionized (one electron removed) atoms are accelerated and then passed thro

Question

Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 153 V/m and the magnetic field is 3.13×102 T . The ions next enter a uniform magnetic field of magnitude 1.73×102 T that is oriented perpendicular to their velocity

A) How fast are the ions moving when they emerge from the velocity selector?

B) If the radius of the path of the ions in the second magnetic field is 17.6 cm , what is their mass?

Explanation / Answer

part A:

to pass the velocity selector without any deflection,

electric force+magnetic force=0

==>q*E=q*v*B (as all the fields and direction of motion are perpendicular to each other)

==>E=v*B

==>v=E/B=153/(3.13*10^(-2))=4888.1789 m/s

part B:

in the second field, magnetic force=centripetal force

==>q*v*B=m*v^2/r

where q=charge on the particle

r=radius of the path=0.176 m

hence m=q*B*r/v=1.6*10^(-19)*1.73*10^(-2)*0.176/4888.1789=9.9662*10^(-26) kg

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