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

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