Question

A particle with mass \(m\) and charge \(q\) is moving within both an electric field and a magnetic field, \(\vec{E}\) and \(\vec{B}\). The particle has velocity \(\vec{v}\), momentum \(\vec{p}\), and kinetic energy, \(K\). Find general expressions for \(d \vec{p} / d t\) and \(d K / d t\), in terms of these seven quantities.

Short answer

Question: Express the rate of change of momentum and the rate of change of kinetic energy of a charged particle in terms of its mass (m), charge (q), electric field (E), magnetic field (B), and velocity (v). Answer: The rate of change of momentum (d𝑝/dt) and the rate of change of kinetic energy (dK/dt) are given by: d𝑝/dt = q(E + v × B) dK/dt = v⋅q(E + v × B)

Step-by-step solution

Calculate the rate of change of momentum#

To find the rate of change of momentum, we will use the Lorentz force equation. The Lorentz force acting on the particle when it is exposed to both electric and magnetic fields is given by: \[ \vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\] Since \(\vec{F} = \dfrac{d\vec{p}}{dt}\), then we can write the expression for \(d\vec{p}/dt\) as follows: \[\frac{d\vec{p}}{dt} = q(\vec{E} + \vec{v} \times \vec{B})\]

Calculate the rate of change of kinetic energy#

To find the rate of change of kinetic energy, we first recall the expression for kinetic energy: \(K = \dfrac{1}{2}mv^2\). Next, we need to find the derivative of \(K\) with respect to time \(t\). We have: \[dK = m\vec{v}\cdot d\vec{v}\] Dividing both sides by \(dt\) gives us: \[\frac{dK}{dt} = m\vec{v}\cdot\frac{d\vec{v}}{dt}\] Recall that force \(\vec{F}\) is related to mass \(m\) and acceleration \(\vec{a} = d\vec{v}/dt\) as \(\vec{F} = m\vec{a}\). From Step 1, we can write \(\vec{F}\) in terms of given quantities. Thus, we can write the rate of change of kinetic energy as follows: \[\frac{dK}{dt} = \vec{v}\cdot q(\vec{E} + \vec{v} \times \vec{B})\] In conclusion, we have expressed the rate of change of momentum and the rate of change of kinetic energy as follows: \[\frac{d\vec{p}}{dt} = q(\vec{E} + \vec{v} \times \vec{B})\] \[\frac{dK}{dt} = \vec{v}\cdot q(\vec{E} + \vec{v} \times \vec{B})\]