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)
9.12 kV
Physics & Space Science
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