A process is accelerated from rest through a potential difference of 1.0 kV. It

Question

A process is accelerated from rest through a potential difference of 1.0 kV. It then enters the velocity selector that has a magnetic field of 4.5 mT into the page. It travels straight through the velocity selector to a region where only the 4.5 mT magnetic field exits. The distance between the plates in the velocity selector in 3 cm. Label the drawing with the appropriate electric charge, electric field, and magnetic field. What is the voltage between the planes of the velocity selector if the distance between the plates is 3 cm? Draw the correct path for the protien. Where does the proton strike the plate? Voltage =

Explanation / Answer

Its pretty difficult to add pictures in answers, so I will describe.

Part A)

The electric charge traveling to the right is labeled as positive, with a charge of 1.6 X 10-19 C

It enters the velocity selector (The horizontal plates) with a magentice field 'B' directed into the page (use x's to represent) and magnitude of 4.5 X 10-3 T

Due to the right hand rule, this field will cause a force on the proton upward. Therefore, to counteract, the Electric Field must point downward. So, draw arrows downward to represent the Electric Field.

Part B)

The proton is accelerated by 1000 V and we can use the conservation of energy to find the velocity

qV = .5mv2

(1.6 X 10-19)(1000) = (.5)(1.67 X 10-27)(v2)

v = 4.38 X 105 m/s

From this velocity, we can find the electric field

v = E/B

E = vB = (4.38 X 105)(4.5 X 10-3) = 1.97 X 103 N/C

Then the voltage is found by

V = Ed = (1.97 X 103)(.03) = 59.1 V

Part C)

The proton leaves the selector and enters the magnetic field to the right (enters between the vertical lines where the gap is). Due to the right hand rule, where the B field is into the page (Again use X to represent). The force will direct the proton to curl in a semicircle upward.

Part D)

The radius of the path is found by

r = mv/qB

r = (1.67 X 10-27)(4.38 X 105)/(1.6 X 10-19)(4.5 X 10-3)

r = 1.0156 m

Therefore, it strikes the plate at twice this ditance (the diameter) above the opening

It strikes 2.03 m above the opening.

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