(adsbygoogle = window.adsbygoogle || []).push({}); 1. An ionized deuteron (a particle with a + e charge) passes through a velocity selector whose perpendicular magnetic and electric fields have magnitudes of 40 mT and 8.0 kV/m, respectively. Find the speed of the ion

2. When a charge particle moves in a velocity selector, both electric field and magnetic field are there.

Any charged particle in an electric field will feel a force proportional to the charge and field strength such that F = qE , where F is force, q is charge, and E is electric field. Similarly, any particle moving in a magnetic field will feel a force proportional to the velocity and charge of the particle. The force felt by any particle is then equal to F = qv X B , where F is force, q is the charge on the particle, v is the velocity of the particle, B is the strength of the magnetic field, and X is the cross product. In the case of a velocity selector, the magnetic field is always at 90 degrees to the velocity and the force is simplified to F = qvB in the direction described be the cross product.

F = qE and F = qvB

So, qE = qvB

V = E/B

Setting the two forces to equal magnitude in opposite directions it can be shown that V = E/B . Which means that any combination of electric (E) and magnetic (B) fields will allowed charged particles with only velocity (v) through.

3. E = 8 KV/m = 8 x 103 V/m

B = 40 mT = 40 x 10-3 T

V = 8 x 103 x 40 x 10-3

V = 320 m/sec

(Ans)

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# Homework Help: Charged Particles in Magnetic Fields

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