-plane, where a uniform magnetic field exists penetrating through the -plane for the left half. In the beginning when the charged particle is inside the magnetic field, the path is horizontal in the direction. After getting out the magnetic field, at the height of , the path becomes a projectile motion, ending on the ground at last. If the magnetic field intensity is , the mass of the charged particle is and the quantity of the electric charge of the particle is then what is the distance above? (Gravitational acceleration is , and the amount of air resistance is negligible.)
The above shows the path (green arrows) of a charged particle in an
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First of all the charged particle is not moving circular in uni. mag. field so the force qvB is balanced by mg thus v=mg/bq. further , this velocity is acting as constant horizontal component in projectile. now for vertical motion of projectile , H=1/2gt*t By 2nd law of kinematics by newton. thus by rearrangement we get 't' and then applying same law to horizontal motion and putting value of t we get this.