AP Physics 2 Exam Question

A long straight wire carries a steady current of 4 A in the positive x-direction. The wire is placed in a magnetic field that points in the positive y-direction. The magnitude of the magnetic field is given by the equation:

B = 2x^2 + 3x + 1 (in Tesla), where x is the distance in meters from the wire, along the y-axis.

a) Determine the direction of the magnetic force that will act on a positive charge moving parallel to the wire and in the positive y-direction.

b) A positive charge of 2 C moves parallel to the wire and at a distance of 1 m from it. Calculate the magnitude and direction of the magnetic force acting on the charge.

c) If the charge were moving at a velocity of 3 m/s, perpendicular to the wire, calculate the magnitude and direction of the magnetic force acting on the charge.

Answer

a) The direction of the magnetic force on a positive charge moving parallel to the wire can be determined using the right-hand rule. According to the right-hand rule, when the thumb points in the direction of the current, the fingers curl in the direction of the magnetic field lines. Therefore, the magnetic force on the positive charge will be in the negative z-direction.

b) To calculate the magnitude and direction of the magnetic force acting on the charge, we can use the formula:

F = qvBsinθ

Where: F is the magnetic force, q is the charge, v is the velocity of the charge, and B is the magnetic field.

Given: q = 2 C v = 0 m/s (since the charge is moving parallel to the wire) B = 2x^2 + 3x + 1 Tesla

At a distance of 1 m from the wire, the magnetic field is: B = 2(1)^2 + 3(1) + 1 B = 2 + 3 + 1 B = 6 Tesla

Substituting these values into the formula, we get:

F = (2 C)(0 m/s)(6 Tesla)sinθ F = 0 N

Therefore, the magnitude of the magnetic force acting on the charge is 0 N. Since the force is 0, the direction does not matter.

c) Since the charge is moving perpendicular to the wire, the angle between the velocity and the magnetic field is 90 degrees. The formula to calculate the magnitude of the magnetic force is:

F = qvBsinθ

Given: q = 2 C v = 3 m/s B = 2x^2 + 3x + 1 Tesla

At a distance of 1 m from the wire, the magnetic field is: B = 2(1)^2 + 3(1) + 1 B = 2 + 3 + 1 B = 6 Tesla

Substituting these values into the formula, we get:

F = (2 C)(3 m/s)(6 Tesla)sin90° F = (2 C)(3 m/s)(6 Tesla)(1) F = 36 N

Therefore, the magnitude of the magnetic force acting on the charge is 36 N, and the direction is determined by the right-hand rule.