Physics · Ch 3 — Magnetism and Magnetic Effects of Electric Current

Motion of a Charged Particle in a Uniform Magnetic Field

3.10.2

Motion of a Charged Particle in a Uniform Magnetic Field

Because $\vec F_m$ is always perpendicular to $\vec v$ , $\vec F_m\cdot\vec v = 0$ at every instant -- the magnetic force does no work and never changes a particle's speed or kinetic energy, only its direction. When a charge $q$ (mass $m$ ) enters a uniform field $\vec B$ with velocity $\vec v$ exactly perpendicular to $\vec B$ , the magnetic force acts as a constant-magnitude centripetal force, bending the particle into a circular path. Setting $qvB = mv^2/r$ gives the radius, period, frequency, and angular frequency of this circular motion:

r = \frac{mv}{qB} = \frac{p}{qB}, \qquad T = \frac{2\pi m}{qB}, \qquad f = \frac{1}{T} = \frac{qB}{2\pi m}, \qquad \omega = \frac{q}{m}B

These last three are together called the cyclotron frequency (or gyro-frequency) relations, and crucially $T$ , $f$ , and $\omega$ depend only on the charge-to-mass ratio $q/m$ -- not on the particle's speed or the radius of its path (a faster particle simply orbits on a proportionally larger circle, in the same time). …

Figure 3.45Circular motion of a charged particle in a perpendicular uniform magnetic field

What this figure shows. A positive charge +q enters a region of uniform field B (drawn into the page) with velocity v perpendicular to B. The Lorentz force F is drawn pointing toward the centre of a circle of radius r, showing the particle continually curving inward under this centripetal magnetic force and so tracing out a full circular path within the field region. …

Figure 3.46Helical path of a charged particle in a uniform magnetic field

What this figure shows. A negative charge -q moves through a uniform field B with its velocity tilted at some angle to B rather than exactly perpendicular to it. The resulting trajectory is drawn as a helix wound around the field lines: the component of velocity along B carries the particle steadily forward along the field direction while the perpendicular component makes it circle around, the two combining into a corkscrew-shaped …

Misc Example 3.19Speed of an electron in circular motion

Worked out. An electron moves perpendicular to a 0.500 T field on a circular path of radius 2.50 mm. From r = mv/(qB), v = qrB/m = (1.60x10^-19)(2.50x10^-3)(0.500)/(9.11x10^-31), which comes out to about 2.20x10^8 m/s -- illustrating how a measured radius, together with the known charge-to-mass ratio of the electron, directly gives its spe …

Misc Example 3.20Helical path of a proton -- radius and pitch

Worked out. A proton moving in a 0.500 T field along x has an initial velocity with both an x-component (along B, unaffected by the magnetic force) and a z-component (perpendicular to B, which curls into a circle). The acceleration at t=0 comes purely from the perpendicular velocity component crossed with B, giving a force and hence acceleration in the y-direction of about 9.58x10^12 m/s^2. Since the velocity is not purely perpendicular to B, the path is helical, not circular: its radius R = m v_perp/(qB) works out to about 4.18 mm, and its pitch (distance advanced along x per full revolution, P = v_x T with T=2 pi m/(qB) approx 13.1x10^-8 s) works out to about 25.5 mm -- roughly six times the radius, because the parallel velo …

Misc Example 3.21Separating uranium isotopes by radius of circular path

Worked out. Two singly-ionized uranium isotopes, mass 3.90x10^-25 kg (U-235) and 3.95x10^-25 kg (U-238), enter a 0.500 T field at 1.00x10^5 m/s. Their radii, r = mv/(qB), come out to r235 approx 48.8 cm and r238 approx 49.4 cm, so after each completes a semicircle the two beams land a distance d = 2r238 - 2r235 approx 1.2 cm apart -- turning a tiny 1.3% mass difference into an easily measurable spatial separation, which is exactly the working principle of a mass spectrometer. The time for each isotope's semicircle, t = (pi r)/v, comes out to about 9.76 microseconds for U-235 a …

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