Section 4.1 · Further exercises · 1 of 20
Exercise 4.1.1: Lorentz force in components
Twenty further exercises in the spirit of the problems, the worked examples and the home exercises, graded from easy to harder within each group. Exercises 4.1.1–4.1.10 belong to the theory of Lecture 8 (the Lorentz force, Biot–Savart's law, flux, force and the vector potential) and Exercises 4.1.11–4.1.20 to the exercise session of Lecture 9 (Ampère's law, coils and rotating charges, measuring and shaping fields). They use only
the two results of the section, for a long straight wire and on the axis of a loop, and superposition.
Lorentz force in components
#cf. the Lorentz force
A charge μC moves with the velocity m/s through a region where kV/m and T. Find (a) , (b) the electric, the magnetic and the total force, (c) the acceleration if kg, (d) the power delivered to the charge by each field.
Straight use of the component form (4.1.4) of the Lorentz force. For the power, remember that a force perpendicular to the velocity delivers none.
With and the component formula gives
so V/m. The -component of , parallel to , contributes nothing.
m/s: the velocity starts to turn in the -plane while is unchanged.
. The magnetic force is perpendicular to and delivers no power. The electric force is perpendicular to at this instant too, so as well: the kinetic energy is momentarily stationary, and only the electric field will change it later, as acquires a -component.
Units: C(m/s)T C (m/s) N/(A m) N. The magnetic force is times the electric one here although is kV/m: a field of T acting on a charge at m/s is equivalent to kV/m. This is why magnetic deflection wins in fast beams, and why the magnetic force does no work however large it is.