A proton moving at 7500 km/s must maintain a cyclotron orbit of less than 25 cm

Question

A proton moving at 7500 km/s must maintain a cyclotron orbit of less than 25 cm radius to stay confined within a nuclear fusion reactor.

A) What minimum magnetic field is necessary to keep the proton confined?

B) If an electron moving into the page is injected into the plasma torus at 7500 km/s by a velocity selector with a 2.0 T magnetic field pointing straight up what is the magnitude and direction of the electric field of the velocity selector?

Stuck on part A and B,. Ithink i am using the wrong formula would really appreciate if you can show me a step by step. thank you.

Explanation / Answer

Given v = 7500 * 103 m/s

radius r = 0.25 m

from the relation   m * v2 / r   =    q * v * B

therefore B = m * v / ( q * r)

B = (1.67 * 10-27 * 7500000) / ( 1.6 8*10-19 * 0.25)

B = 0.3 T

b)

the electric field

E = v * B

E = 7500000 * 2

E = 15 M N/c

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