Chapter 21, Problem 27 A positively charged particle of mass 6.07 x 10-8 kg is t

Question

Chapter 21, Problem 27 A positively charged particle of mass 6.07 x 10-8 kg is traveling due east with a speed of 86.9 m/s and enters a 0.262-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 4.92 × 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 B (out of screen) 1 (a)Number Units (b) Number Units

Explanation / Answer

(A) d = v t

pi r / 2 = v t

pi r / 2 = (86.9)(4.92 x 10^-3)

r = 0.272 m

Fb = m a_c = m v^2 / r

Fb = (6.07 x 10^-8) (86.9^2) / 0.272

Fb = 1.684 x 10^-3 N .....Ans

(B) Fb = q v B

1.684 x 10^-3 = (q) (86.9) (0.262)

q = 74 x 10^-6 C Or 74 uC

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