Question

A beam of protons is accelerated easterly from rest through a potential difference of 5.0kV . It enters a region where there exists an upward pointing uniform electric field. This field is created by two parallel plates separated by 15cm with a potential difference of 250 V across them.

PART A

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

Express your answer using two significant figures.

PART B

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

Express your answer using two significant figures.

PART C

What is the direction of this magnetic field?

What is the direction of this magnetic field?

toward the north

toward the south

PART D

What happens to the protons if the magnetic field is greater than the value found in part B?

What happens to the protons if the magnetic field is greater than the value found in part B?

	the protons will be deflected downward
	the protons will be deflected upward

Answer 1

m = 1.67 * 10^-27 kg, q = 1.6 * 10^-19 C, V₁ = 5.0 kV, d = 15 cm, V₂ = 250 V

mv² = qV₁

v = sqrt (2qV₁/m) = sqrt (2 * 1.6 * 10-19 * 5 * 10³/1.67 * 10^-27 )

v = 9.79 * 10⁵ m/s

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E = V₂ / d = 250 / 15 * 10-2

= 1666.67 V/m

B = E/v = 1666.67 / 9.79 * 10⁵

B = 1.70 * 10-3 = 1.70 mT

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towards south

Using Flemings left hand rule direction of the field should be towards south.

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the protons will be deflected downward