# Electromagnetism

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A particle with a mass of 6.64x10^-27 kg and a charge of +3.2x10^-19 C is accelerated from rest through a potential difference of 2.45x10^6 V. the particle then enters a uniform magnetic field of strength 1.60 T.

a) Find the particle's gain in KE as it moves through the potential difference

b) What is the velocity of the particle as it enters the magnetic field?

c) If the particle's velocity is perpendicular to the magnetic field at all times, what is the magnitude of the magnetic force exerted on the particle as it passes through the magnetic field.

• Electromagnetism -

KE=mv²/2= qU
v=sqrt(2qU/m)=…
F=qvB=…

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