1. A positively charged particle of mass 7.28×10 -8 kg is traveling due east wit

Question

1. A positively charged particle of mass 7.28×10-8kg is traveling due east with a speed of 89.1m/s and enters a 0.300T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 2.11×10-3s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. What is the magnitude of the magnetic force acting on the particle?

2. A 3.12mF (microF) and a 5.20mF (microF) capacitor are connected in series across a 27.0V battery. A 7.09mF (microF) capacitor is then connected in parallel across the 3.12mF (microF) capacitor. Determine the voltage across the 7.09mF (microF) capacitor.

Explanation / Answer

1) T = 4 * t = 4 * 2.11 * 10-3 = 8.44 * 10-3 s

v = 2 pi r / T

r = v T / 2 pi = (89.1 * 8.44 * 10-3) / (2 pi)

r = 0.12 m

F = m v2 / r = (7.28 * 10-8 * 89.12) / (0.12)

magnitude of the magnetic force  = 4.82 * 10-3 N

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