Answer to Question #117612 in Electric Circuits for Bio

Question

The Bainbridge-type mass spectrometer uses a velocity selector to select only those ions with the
proper velocity. The selector has two charged parallel plates to create an electric field pointing down,
and copper coils to create a magnetic field. Electrons (q=-1.6x10-19) with speed 5x104 m/s should
pass through the selector with velocities directed to the right without changing direction.
a) In which direction should the magnetic field pointing to?
b) How large should be the magnetic field if the electric field strength is 7000 N/C.
c) What would happen to an electron that has the speed of 7.5x104 m/s?

Answer 1

The electric field and magnetic field are so adjusted that the deflection produced by one field is nullified by the other, so that the ions do not suffer any deflection within the velocity selector. Let E and B be the electric field intensity and magnetic induction respectively and q be the charge of the positive ion. The force exerted by the electric field is equal to qE and the force exerted by the magnetic field is equal to Bqv where v is the velocity of the positive ion.

$qE=Bqv$

$v=E/B$

$5\times10^4=7000/B\\B=0.14 T$

And as the both forces cancel each other so the electron will travel undeflected