what magnetic force would an electron experience due to Earth’s magnetic field if it were moving at 3x10E6 m/s directly away from earth’s surface? (Earth’s B-field in the region is parallel to the surface and has a strength of 550 µT)
To calculate the magnetic force experienced by an electron due to Earth's magnetic field, you can use the equation:
F = q*v*B*sin(θ)
Where:
F is the magnetic force,
q is the charge of the electron (-1.6 x 10^-19 C),
v is the velocity of the electron (3 x 10^6 m/s),
B is the strength of the magnetic field (550 µT or 550 x 10^-6 T),
θ is the angle between the velocity vector and the magnetic field vector.
In this case, since the electron is moving directly away from Earth's surface, the angle between the velocity vector and the magnetic field vector is 90 degrees, so sin(θ) = 1.
Now let's plug in the values into the equation:
F = (-1.6 x 10^-19 C) * (3 x 10^6 m/s) * (550 x 10^-6 T) * 1
Simplifying the equation:
F = -8.8 x 10^-13 N
Therefore, the magnetic force experienced by the electron due to Earth's magnetic field would be -8.8 x 10^-13 Newtons. Note that the negative sign indicates that the force acts in the opposite direction to the electron's velocity.