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? (m_proton = 1.67 × 10^-27 kg,e = 1.60 × 10^-19
C)

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The Answer of 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? (m_proton = 1.67 × 10^-27 kg,e = 1.60 × 10^-19 C)

A Proton Is First Accelerated from Rest Through a Potential