Question

A charged particle ( mass = 4.0 µg, charge = 5.0 µC ) moves in a region where the only force on it is magnetic. What is the magnitude of the acceleration of the particle at a point where the speed of the particle is 5.0 km/s, the magnitude of the magnetic field is 8.0 mT, and the angle between the direction of the magnetic field and the velocity of the particle is 60°?

Answer 1

Let us determine the magnetic force. It is Lorentz force equal to $|F| = |q\vec{v}\times\vec{B}| = q\cdot v\cdot B\cdot \sin \alpha,$ where $\alpha$ is the angle between the direction of the magnetic field and the velocity of the particle. Therefore,

$F = q\cdot v\cdot B\cdot \sin \alpha, \\ F = 5.0\cdot10^{-6}\,\mathrm{C}\cdot 5.0\cdot10^3\,\mathrm{m/s}\cdot 8.0\cdot10^{-3}\,\mathrm{T}\cdot \sin60^\circ,\\ F = 1.7\cdot10^{-4}\,\mathrm{N}.$

The acceleration can be obtained as $a = \dfrac{F}{m} = \dfrac{1.7\cdot10^{-4}\,\mathrm{N}}{4.0\cdot10^{-6}\cdot10^{-3}\,\mathrm{kg}} \approx 4.3\cdot10^4\,\mathrm{m/s^2}.$