Consider a charged particle moving through a region in which the electric field

Question

Consider a charged particle moving through a region in which the electric field is perpendicular to the magnetic field, with both fields perpendicular to the initial velocity of the particle (see figure below). Such a device is called a velocity selector because for one particular value of the particle speed vselect the particle will travel straight through the device without being deflected, whereas particles with speeds that are either less than or greater than vselect will be deflected.

a) What is the direction of the electric and magnetic forces on the ions moving through the selector?

b) what is the magnitude of the electric and magnetic forces on the ions moving through the selector?

c? Find an expression for Vselect as a function of E, B, g, q (charge of the ion) and m (mass of the ion) as appropriate.

d) Estimate this value for a proton (m_p_ = 1.7 e -27 kg, q_p_ = e)

Explanation / Answer

A.) The direction of electric force will be in the direction of electric field. The direction of magnetic field will be perpendicular to the velocity of the ion and also perpendicular to that magnetic filed.

B.)

The force exerted on a charged particle by the electric field is given by:

                                          F = qE

The magnitude of the force exerted by the magnetic field is F = qvB, as long as the velocity is perpendicular to the field.

C.) The force due to E is F= q E where q= charge
The force due to B is F=qvB where v= velocity

For there to be no deflection, these must be equal qE=qvB i.e. E=vB

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