# Physics help..........?

A positively charged particle of mass 6.69 × 10-8 kg is traveling due east with a speed of 82.2 m/s and enters a 0.381-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 1.27 × 10-3 s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.

### 1 Answer

- NCSLv 72 months ago
distance traveled

s = v*t = 82.2m/s * 1.27e-3s = 0.104 m

and

r = s / Θ = 0.104m / π/2 = 0.0665 m

radial acceleration

a = v²/r = (82.2m/s)² / 0.0665m = 1.02e5 m/s²

F = ma = 6.69e-8kg * 1.02e5m/s² = 6.80e-3 N ◄ force

F = qvB

6.80e-3 N = q * 82.2m/s * 0.381T

q = 2.17e-4 C ◄ charge

or 217 µC

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