A beam of protons is accelerated easterly from rest through a potential differen

Question

A beam of protons is accelerated easterly from rest through a potential difference of 3.0kV . It enters a region where there exists an upward pointing uniform electric field. This field is created by two parallel plates separated by 17cm with a potential difference of 250 V across them.

Part A: What is the speed of the protons as they enter the electric field?

Part B: Find the magnitude of the magnetic field (perpendicular to E? ) needed so the beam passes undeflected through the plates.

Part C: What is the direction of this magnetic field? toward the north or south?

Part D: What happens to the protons if the magnetic field is greater than the value found in part B? The protons will be deflected upward or the protons will be deflected downward?

Explanation / Answer

a)

By Conservation of energy

(1/2)mv2=qV

(1/2)*(1.67*10^-27)*v2=(1.6*10^-19)*3000

v=7.48*105 m/s

b)

Electric field

E=V/d =250/0.17=1470.6 N/C

Magnetic field

B=E/v=1470.6/(7.58*10^5)

B=1.94*10^-3 T or 1.94 mT

c)

Direction of magnetic field is toward South

d)

the protons will be deflected downward

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