Magnetic force calculations sometimes cause confusion for students because the calculation is different depending on the polarity of the moving particle (that is, whether the particle is a negatively charged electron or positively charged proton). For example:
Suppose a proton is moving at 8.0 · 10^6 m/s east, perpendicularly through an applied magnetic field (directed north) of 1.0 T. What is the magnitude of the magnetic force exerted on the proton, and in what direction? If an electron was moving through the same field at the same angle with the same velocity, what would be the direction of the magnetic force?
To begin with, physics students will recall that
Suppose a proton is moving at 8.0 · 10^6 m/s east, perpendicularly through an applied magnetic field (directed north) of 1.0 T. What is the magnitude of the magnetic force exerted on the proton, and in what direction? If an electron was moving through the same field at the same angle with the same velocity, what would be the direction of the magnetic force?
To begin with, physics students will recall that
F(magnetic) = qvB(sine θ)
where q is the charge of the proton, v is its velocity, and B is the magnitude of the magnetic field. Since the proton moves perpendicularly through the field, sine θ = 1. Substituting, we see
F(magnetic) = (1.6 · 10^-19 C)(8.0 · 10^6 m/s)(1.0 T) = 1.3 · 10^-12 N
Now that we have calculated the magnitude, what is the direction? Use the right hand rule. With flat hand, point your fingers in the direction of the particle’s motion (east). Curl your fingers towards the direction of the magnetic field (north). Your thumb points in the direction of the magnetic force (upwards), perpendicular to the xy-plane. For an electron, everything would be the same except the magnetic force would point downwards but perpendicular to the xy-plane (see http://www.youtube.com/watch?v=LK7hv4LX3ys for a helpful video).
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