Exiting from a magnetic field

The above shows the path (green arrows) of a charged particle in an x y xy -plane, where a uniform magnetic field exists penetrating through the x y xy -plane for the left half. In the beginning when the charged particle is inside the magnetic field, the path is horizontal in the + x +x direction. After getting out the magnetic field, at the height of H H , the path becomes a projectile motion, ending on the ground at last. If the magnetic field intensity is B B , the mass of the charged particle is m , m, and the quantity of the electric charge of the particle is q , q, then what is the distance L L above? (Gravitational acceleration is g g , and the amount of air resistance is negligible.)

m 2 g H q B \frac{m\sqrt{2gH}}{qB} m g H q B \frac{m\sqrt{gH}}{qB} m 2 q H g B \frac{m\sqrt{2qH}}{gB} 2 g H q B \frac{\sqrt{2gH}}{qB}

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1 solution

Akash Tandale
Jun 27, 2014

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.

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