A point particle q that has charge = 4.5 x 10^-6 C and mass 10^-5 Kg, moves at a

Question

A point particle q that has charge = 4.5 x 10^-6 C and mass 10^-5 Kg, moves at a velocity 3 m/s x along the line y = 3m in the x-y plane as shown. A magnetic field = 1 Tesla is turned on in the y direction, B=1 z Assume at time t=0, the charge q is at x=0, y = 3 m and z = 0. a) What is the direction of the force on the particle? b) What is the magnitude of the force on the particle? c) What is the acceleration of the particle in the direction of the force? d) Knowing that the final z position of the particle, assuming that the acceleration is in the z direction is z(t)=1/2 at^2 where a is the acceleration, what are the coordinates of the particle at time t = 1 second?

Explanation / Answer

Part A)

By the right hand rule, the force is in the z direction

Part B)

F = qvB

F = (4.5 X 10-6)(3)(1)

F = 1.35 X 10-5 N

Part C)

F = ma

(1.35 X 10-5) = (1 X 10-5)a

a = 1.35 m/s2

Part D)

Constant velocity in the x moves it as per d = vt

d = 3(1) = 3

No motion in the y, it will stay at 3

In the z...

d = .5(1.35)(12)

d = .675

Thus the coordinates are (3, 3, .675)

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