A positively charged particle of mass 6.20 10-8 kg is traveling due east with a

Question

A positively charged particle of mass 6.20 10-8 kg is traveling due east with a speed of 30 m/s and enters a 0.43-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 6.20 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? Incorrect: Your answer is incorrect. N (b) Determine the magnitude of its charge.

Explanation / Answer

r = distance traveled/(pi/2)

= (vt/(pi/2))

= 2vt/

=(2)(30)(6.2*10^-3)/3.14

= 0.118 m

Force = mv^2/r

= 6.2*10^-8 * 30^2 /0.118

= 4.72*10^-4 N

The force is

F = qvB

Hence the charge is

q = F/vB = (4.72*10^-4)/(30)(0.43)=6.77*10^-6 C

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