Charged Particles in Magnetic Fields

[predentalgirl1]

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)

 

[Shooting Star]

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)

Why have you multiplied E and B?

(I really don't feel that you should copy and paste almost the entire page from Wikipedia, along with the grammatical and spelling mistakes. Write it yourself.)

 

[predentalgirl1]

Why have you multiplied E and B?

(I really don't feel that you should copy and paste almost the entire page from Wikipedia, along with the grammatical and spelling mistakes. Write it yourself.)

If I should not multiply E and B, what
should I do instead to get the correct answer?

 

[Shooting Star]

F = qE and F = qvB

So, qE = qvB

V = E/B

This is from your post. What do you think you should do to get v?

 

[predentalgirl1]

Given that B =40 x 10 ^-6
E = 8000 V/m

have Bev = eE

v = E/B

v= 8000/ 40 x 10^-6
v= 2 x 10^8m/s

 

[Shooting Star]

Given that B =40 x 10 ^-6

Is the value of B you have used here is correct?