Question

A particle of charge q is fixed at the origin of an xy coordinate system. At t = 0 a particle m = 0.778 g, q = 5.07 µC is located on the x axis at x = 15.1 cm, moving with a speed of 58.0 m/s in the positive y direction. For what value of Q will the moving particle execute circular motion?

Answer 1

To find the value of Q for which the moving particle executes circular motion, we can use the equation for the centripetal force. After substituting the given values and solving, we find that Q is approximately 0.0136 C.

Step-by-step explanation:

To find the value of Q for which the moving particle executes circular motion, we can use the equation for the centripetal force:

F = λrv²

where F is the force, λr is the magnitude of the magnetic field, and v is the velocity. In circular motion, the magnetic force provides the necessary centripetal force. The magnetic force is given by:

F = |q|vB

where |q| is the magnitude of the charge, v is the velocity, and B is the magnetic field. Equating the two expressions for F, we have:

|q|vB = λrv²

Simplifying, we find:

|q|B = λrv

Substituting the given values:

(5.07 µC)(Q) = (0.778 g)(0.58 m/s)(0.151 m)

Simplifying further:

Q = (0.778 g)(0.58 m/s)(0.151 m)/(5.07 µC) ≈ 0.0136 C