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
Question
Answer:
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