Motion in Electromagnetic Fields

Uniform Fields

A uniform field ($\bm{F}$) is a field with spatial invariance, therefore, a zero gradient ($\nabla \bm{F} = 0$).

Test Charge

A test charge is an idealized particle where all other properties are assumed to be negligible except for the charge.

Lorentz Force Law

Coulomb's Law is only valid for electrostatics, as magnetic force is also induced by moving charge. The combined force ($\bm{F}$) on the test charge ($q$) can be governed by Lorentz force law:

$$\begin{aligned} \bm{F} = q(\bm{E}+\bm{v} \times \bm{B}) \end{aligned}$$

When only considering the magnitude of the force:

$$\begin{aligned} \bm{F} = q(|\bm{E}| + |\bm{v} \times \bm{B}|) = q(|\bm{E}|+|\bm{v}||\bm{B}|sin\theta) \end{aligned}$$

The magnitude form is written in the IB formula booklet separately for magnetic and electric force:

$$\begin{aligned} F &= qvB \sin \theta\\ F &= qE\\ E &= \frac{F}{q} \end{aligned}$$

Motion of Current in Electromagnetic Field

The total force ($\bm{F}$) on a current ($I$) is determined by the combine force on the total number ($N$) of moving charges ($q$) in the current. As the dimensions of a electron is negligible when comparing to the cross section of a wire, the flow of electrons $I$ can be approximated by a continuum of current density ($\bm{J} = nq\bm{v}$), where $n$ is the density of charge carriers, and since the electric field have a negligible effect, only the force applied by magnetic field ($\bm{B}$ is accounted):

$$\begin{aligned} \bm{F_m} &= nV(q(v\times \bm{B})) = V(\bm{J}\times\bm{B})\\ \bm{F_m} &= \int_C \bm{J}\times\bm{B} dV \end{aligned}$$

For current of a conductor with constant cross-sectional area ($A$) and length displacement of ($\bm{L}$), the volume can be evaluated as ($V = AL$), leading to a force of:

$$\begin{aligned} \bm{F_m} = \int I(d\bm{L})\times B = I \int \bm{B}\times d\bm{L} \end{aligned}$$

Since IB only considers interactions of current with uniform electric fields, magnetic field $\bm{B}$ is constant, evaluating the force to:

$$\begin{aligned} \bm{F_m} = I(\bm{L}\times\bm{B}) \end{aligned}$$

Since IB considers the direction separately by the use of right hand rule (Motor effect), the magnitude of the force is expressed as:

$$\begin{aligned} |\bm{F_m}| = F = I|\bm{L}\times\bm{B}| = I|\bm{L}||\bm{B}|\sin\theta = BIL\sin\theta \end{aligned}$$

This is expressed in the formula booklet $D.3$.

Motion Between Currents

According to Ampere's Law, current ($I_1$) generates a magnetic field ($\bm{B_1}$), due to rotational symmetry around the direction of current, the magnetic field can be express using cylindrical coordinates ($\hat{\phi}$) shows:

$$\begin{aligned} \bm{B_1} = \frac{\mu_0 I_1}{2\pi r}\hat{\bm{\phi}} \end{aligned}$$

For a $\hat{\phi} = \hat{y}$ where $\hat{y}$ is a unit vector perpendicular to other current ($I_2$). Using the Lorentz Force Law, the magnetic force ($F_{12}$) experienced by the other current is:

$$\begin{aligned} \bm{F_{12}} &= I_2(\bm{L}\times\bm{B_1})\\ \bm{F_{12}} &= I_2L(\hat{z}\times{B_1}), \quad \hat{z} = \hat{\bm{L}}\\ \bm{F_{12}} &= I_2L\left(\hat{z}\times\frac{\mu_0 I_1}{2\pi r}\hat{y}\right)\\ \bm{F_{12}} &= -\frac{\mu_0 I_1 I_2 L}{2\pi r}\hat{x}, \quad \hat{y} \times \hat{z} = -\hat{x} \end{aligned}$$

Similar to before the IB expresses the magnitude form of this equation in the formula booklet:

$$\begin{aligned} |\bm{F_{12}}| = F = -\frac{\mu_0 I_1 I_2 L}{2\pi r}\hat{x} = \mu_0 \frac{I_1 I_2 L}{2\pi r} \end{aligned}$$

Where the direction of motion on current can be determined by right hand rule (motor effect).