Magnetic Fields! Physics fun!?

Question

A 5.3×10−9 kg particle carrying a charge of 3.2×10−6 C is accelerated by a potential difference of 3,200 V from rest. It then passes into a uniform magnetic field of strength 4.5×10−3 T. (a) What is the largest force it can experience? (b) What is the smallest force it can experience?

Answer 1

The final velocity of the particle after being accelerated through the potential difference is:

v = (2q∆V / m)^(1/2)

= ((2 x (3.2 x 10^-6 C) x 3200 V) / (5.3 x 10^-9 C))^(1/2)

= 2,000 m/s

The maximum force acts on the particle when it moves perpendicular to the direction of the field. The minimum force is zero when the particle travels parallel to the direction of the field.

F = qvB

= (5.3 x 10^-9 C) x 2,000 m/s x (4.5 x 10^-3 T)

= 4.8 x 10^-8 N