The electric field between the plates of the velocity selector is 950 V/m and th

Question

The electric field between the plates of the velocity selector is 950 V/m and the magnetic field in both parts of the apparatus is 0.930 T. Calculate the radius of the path for a singly charged ion with mass m= 2.18 x 10^-26 kg.
Please show all steps The electric field between the plates of the velocity selector is 950 V/m and the magnetic field in both parts of the apparatus is 0.930 T. Calculate the radius of the path for a singly charged ion with mass m= 2.18 x 10^-26 kg.
Please show all steps
Please show all steps

Explanation / Answer

The velocity selector works by exerting electric force qE and magnetic force(oppositely directed) qVB on the ion of charge q. When the forces are equal the ion travels undeflected and enters the 2nd chamber.
When the forces are equal;

qVB = qE so V = E/B, the velocity of the ion entering 2nd chamber.

In 2nd chamber the magnetic force , qVB , causes circular motion so it is a centripital force and from 2nd law;

F = ma

qVB = mV^2/R

R = mV/qB

Using V found above;

R = mE/qB^2

So R = m*E/(q*B^2) = 2.18x10^-26*950/(1.60x10^-19*(0.930)^2=1.49*10^-4m

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