A long wire lies in the page and along the vertical direction. The current I flo

Question

A long wire lies in the page and along the vertical direction. The current I flows upward vertically along the page. A positive charge has speed v and moves in a direction opposite the current in the wire. The value of the charge is q. The perpendicular distance between the charge and wire is d. For most of the parts you will be using only I, q, d, v or other symbols. At the location of the charge, what is the magnitude B of the magnetic field due to the wire? Use symbols. At the location of the charge, what is the direction of the magnetic field due to the wire? Carefully write symbols and/or words at the location of the charge indicating the direction of the magnetic field. If needed sketch another diagram. What is the direction of the magnetic force on the charge? Carefully write symbols and/or words at the location of the charge indicating the direction of the force. If needed sketch another diagram. What is the magnitude F of the magnetic force on the charge? Use symbols. Evaluate the magnitude F of the magnetic force if I = 2.00(A), d = 0.10 m, q = 1.60 times 10-19 C (the magnitude of the electron charge) and speed v = 2.00 times l06 m/s. Assume mu o = 4 pi times 10-7 T m/A.

Explanation / Answer

Part A)

The magnetic field at the location of the charge is

B = oI/2d

Part B)

Since we don't know that actual location of the charge, and the orientation of its location with respect from the wire, we can only say that the magnetic field will act perpendicular to the direction of the travel of the charge, and to the direction of the separation distance.

To draw a picture, consider the wire to the left of the charge and in the same xy plance. The charge is a distance d away in the positive x direction. The current in the wire flow in the positive y direction. The magnetic field will be in the negative z direction.

Part C)

Since the current flows upward, and the charge flows downward, the forces is repulsive. Therefore the force will be in a path directed away from the wire in the direction of the separation distance d.

Using the picture described in part B, that force would be in the positive x direction

Part D)

F = qvB   and B = oI/2d

So F = qvoI/2d

Part E)

F = (1.6 X 10-19)(2 X 106)(4 X 10-7)(2)/(2)(.10)

F = 1.28 X 10-18 N

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