A charged particle q moves at constant velocity through a crossed electric and m

Question

A charged particle q moves at constant velocity through a crossed electric and magnetic fields (E and B, which are both constant in magnitude and direction). Write the magnitude of the electric force on the particle in terms of the variables given. Do the same for the magnetic force. Show that when the forces on the particle (electric and magnetic) balance it means that the particle can only have one specific velocity. What is that velocity? (Your answer will be an algebraic expression, not a number).

Explanation / Answer

Electric force Fe = E q

magnetic force Fm = qv B

Since EF and magnetic fields are Perpendicular

Fe = Fm

Eq = q v B

E = v B   is the relation between EF and MF in Crosse Fields

then the equation of Velocity is

V = E/B

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