QuestionJune 5, 2025

Ch 22 # 8 Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500mu C charge and flies due west at a speed of 660m/s over the Earth's magnetic south pole (near Earth's geographic north pole)where the 8.00times 10^-5T magnetic field points straight down. What are the direction and the magnitude of the magnetic force on the plane?

Ch 22 # 8 Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500mu C charge and flies due west at a speed of 660m/s over the Earth's magnetic south pole (near Earth's geographic north pole)where the 8.00times 10^-5T magnetic field points straight down. What are the direction and the magnitude of the magnetic force on the plane?
Ch 22 # 8 Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500mu C charge and flies due west at a speed of 660m/s over the Earth's magnetic south pole (near Earth's geographic north pole)where the 8.00times 10^-5T magnetic field points straight down. What are the direction and the magnitude of the magnetic force on the plane?

Solution
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Answer

Magnitude: 2.64 \times 10^{-5} \, \mathrm{N}; Direction: North Explanation 1. Calculate the magnetic force Use **F = qvB \sin(\theta)**. Here, q = 0.500 \times 10^{-6} \, \mathrm{C}, v = 660 \, \mathrm{m/s}, B = 8.00 \times 10^{-5} \, \mathrm{T}, and \theta = 90^\circ (since velocity is perpendicular to the magnetic field). Thus, F = (0.500 \times 10^{-6}) \times 660 \times 8.00 \times 10^{-5} \times \sin(90^\circ) = 2.64 \times 10^{-5} \, \mathrm{N}. 2. Determine the direction of the force Apply the right-hand rule. The charge is positive, moving west, and the magnetic field points down. The force direction is north.

Explanation

1. Calculate the magnetic force<br /> Use **$F = qvB \sin(\theta)$**. Here, $q = 0.500 \times 10^{-6} \, \mathrm{C}$, $v = 660 \, \mathrm{m/s}$, $B = 8.00 \times 10^{-5} \, \mathrm{T}$, and $\theta = 90^\circ$ (since velocity is perpendicular to the magnetic field). Thus, $F = (0.500 \times 10^{-6}) \times 660 \times 8.00 \times 10^{-5} \times \sin(90^\circ) = 2.64 \times 10^{-5} \, \mathrm{N}$.<br /><br />2. Determine the direction of the force<br /> Apply the right-hand rule. The charge is positive, moving west, and the magnetic field points down. The force direction is north.
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