Question

Aircraft sometimes acquire small static charges. If a supersonic jet has a 0.500-μC charge and flies due west at a speed of 660 m/s over the Earth’s magnetic south pole, where the 8.00×10^(-5)-T magnetic field points straight down, what are the direction and the magnitude of the magnetic force on the plane?

A) 2.64 × 10^(-2) N upward
B) 3.96 × 10^(-2) N downward
C) 1.32 × 10^(-2) N downward
D) 1.98 × 10^(-2) N upward

Answer 1

The magnetic force on the supersonic jet with a 0.500-µC charge flying due west at 660 m/s over the Earth's magnetic south pole is 2.64 × 10-2 N directed upward. The calculation uses the formula F = qvB sin(θ) with a 90-degree angle between the velocity and magnetic field.

Step-by-step explanation:

To determine the direction and magnitude of the magnetic force on a plane with a static charge flying over the Earth's magnetic south pole, we use the formula for the magnetic force on a moving charge (Lorentz force): F = qvB sin(θ), where q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. In this case, the charge q is 0.500 µC (or 0.500 × 10-6 C), the velocity v is 660 m/s, and the magnetic field B is 8.00 × 10-5 T. The angle θ is 90 degrees because the plane flies due west while the magnetic field points straight down.

By substituting the values, we get F = (0.500 × 10-6 C) × (660 m/s) × (8.00 × 10-5 T). This calculation yields a force of 2.64 × 10-2 N. According to the right-hand rule, the force is directed upward for a positively charged object, hence the correct answer is (A) 2.64 × 10-2 N upward.

Part (b) would entail considering whether this force is significant compared to other forces acting on the jet, such as gravity and aerodynamic forces. If the force is much smaller than these, it may be considered negligible.