Question

If a particle of charge q is moving with velocity v along the x-axis and the magnetic field B is acting along the y-axis, use the expression vector F = q( vector v x vector B) to find the direction of the force F acting on it. 

A beam of proton passes undeflected with a horizontal velocity v, through a region of electric and magnetic fields, mutually perpendicular to each other and normal to the direction of the beam. If the magnitudes of the electric and magnetic fields are 50kV/m, 100mT respectively, calculate 

(i) velocity v of the beam. 

(ii) force with which it strikes a target on the screen, if the proton beam current is equal to 0.80mA.

Answer 1

That is, force is acting along z-axis. 

(i) For a beam of charged particles to pass undeflected crossed electric and magnetic fields, the condition is that electric and magnetic forces on the beam must be equal and opposite i.e.,

(ii) The beam strikes the target with a constant velocity, so force exerted on the target is zero. However, if proton beam comes to rest, it exerts a force on the target, equal to rate of change of linear momentum of the beam i.e.,

where n is the number of protons striking the target per second.