Physics Homework question 10

Physics
Tutor: None Selected Time limit: 1 Day

A proton is first accelerated from rest through a potential difference 

V and then enters a uniform 0.750-T magnetic field oriented perpendicular to its path. In this field, the proton follows a circular arc having a radius of curvature of 1.84 cm. What was the potential difference V? (mproton = 1.67 × 10-27 kg, e = 1.60 × 10-19 C)

answer in kV

May 12th, 2015

B = mv/rq
.750 = 1.67*10^-27 * V/.0184*1.6*10^-19
V = 1322155.68 m/sec.
Ke = q*Voltage = 1/2 * m*V^2

potential difference = 9.1kV

May 12th, 2015

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