## A beam of protons is accelerated through a potential difference of 0.750 kVkV and then enters a uniform magnetic field traveling perpendicul

Question

A beam of protons is accelerated through a potential difference of 0.750 kVkV and then enters a uniform magnetic field traveling perpendicular to the field. You may want to review (Pages 641 – 643) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of electron motion in a microwave oven. Part A What magnitude of field is needed to bend these protons in a circular arc of diameter 1.80 mm ? Express your answer in tesla to three significant figures. BpBp = nothing TT SubmitRequest Answer Part B What magnetic field would be needed to produce a path with the same diameter if the particles were electrons having the same speed as the protons? Express your answer in tesla to three significant figures. BeBe = nothing TT SubmitRequest Answer

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6 months 2021-09-04T21:55:43+00:00 1 Answers 3 views 0

Explanation:

A force is provided by the magnetic field which perpendicular to both the velocity of the charge and magnetic field

F = qvB  where B is the magnetic field in tesla, q is charge and v is velocity

The potential energy is transferred into kinetic energy

PE = Vq = 1/2 mv²

v = √(2Vq/ m)

charge on the proton = 1.602 × 10 ⁻¹⁹ C and mass of a proton = 1.673 × 10⁻²⁷

v = √ (( 2 × 1.602 × 10 ⁻¹⁹ C × 0.75 × 10³V ) / (1.673 × 10⁻²⁷)) = √( 1.4363 × 10¹¹ ) = 3.79 × 10⁵ m/s

The force of magnetic field produces centripetal force

qvB = mv² /R

where  R radius = 1.80mm = 0.0018 m / 2 = 0.0009 m

qvB = ( m / R ) × (2qV /m)

cancel the common terms

vB = 2V / R

3.79 × 10⁵ m/s × B = 2 × 0.75 × 10³V / 0.0009 = 1.667 × 10⁶

B = 1.667 × 10⁶ / 3.79 × 10⁵ m/s = 4.40 T

b) magnetic field needed for the electron

qvB = mv² /R where m is the mass of an electron = 9.11 × 10⁻³¹ Kg

qB = mv/R

qB = ( 9.11 × 10⁻³¹ Kg × 3.79 × 10⁵ m/s) / 0.0009

qB  = 3.8363 × 10 ⁻²²

B = 3.8363 × 10 ⁻²² / 1.602 × 10⁻¹⁹ kg = 0.0239 T