Magnetic Force

The magnetic force on a moving charge

The experiment with two hanging current carrying wires shows that a moving charge in a magnetic field feels a force, and that the directions of $\vec{v}$, $\vec{B}$ and $\vec{F}$ are related as a vector cross product.

For an electron of charge $-e$ moving with velocity $\vec{v}$ in a field $\vec{B}$:

$$\vec{F}_m = -e\,(\vec{v}\times\vec{B})$$

and in general, for a charge $q$:

$$\vec{F}_m = q\,(\vec{v}\times\vec{B})$$

Lorentz force

If an electric field $\vec{E}$ and a magnetic field $\vec{B}$ are both present, the net force on a charge $q$ moving with velocity $\vec{v}$ is

$$\vec{F} = q\left[\vec{E} + (\vec{v}\times\vec{B})\right] = q\vec{E} + q(\vec{v}\times\vec{B}) = \vec{F}_e + \vec{F}_m$$

This is the Lorentz force law. $\vec{F}_e$ is the electric part and $\vec{F}_m$ the magnetic part. The two hanging-wire experiments are the experimental justification for it.

Direction, and the diagram

Diagram (Fig. 10.4): draw $\vec{v}$ and $\vec{B}$ as two arrows from a common point, with an angle $\theta$ between them; the plane they define is shaded. The cross product $\vec{v}\times\vec{B}$ is drawn as an arrow perpendicular to that plane, along the unit vector $\hat{n}$ given by the right-hand screw rule (turn $\vec{v}$ into $\vec{B}$; the advance of the screw is $\hat{n}$). For a positive charge $\vec{F}_m$ is along $\hat{n}$; for a negative charge it is opposite to $\hat{n}$.

Consequences of the Lorentz force law

• If $\vec{v}$ is parallel (or antiparallel) to $\vec{B}$, then $\sin\theta = 0$ and the magnetic force is zero.
• If the charge is stationary, $\vec{v} = 0$, the magnetic force is zero even though $\vec{B} \neq 0$.
• $\vec{F}_m$ is always perpendicular to $\vec{v}$, so $\vec{F}_m \cdot \vec{v} = 0$ for any $\vec{B}$. The magnetic force is perpendicular to the displacement, so it does no work on a moving charge.
• Hence a magnetic field may change the direction of motion of a charged particle, but it can never change its speed or its kinetic energy.
• Contrast with the electric force: $\vec{F}_e$ depends only on $\vec{E}$ and $q$; $\vec{F}_m$ depends on $q$, on the magnitude and direction of $\vec{v}$, and on $\vec{B}$ through the cross product.

Unit of magnetic field

From the magnitude of the force,

$$F = q\,\lvert \vec{v}\times\vec{B} \rvert = qvB\sin\theta$$

where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. Rearranging, $B = \dfrac{F}{qv}$, so dimensionally $[B] = [F/qv]$ and the unit of $B$ is N s C$^{-1}$ m$^{-1}$.

If a force of 1 N acts on a charge of 1 C moving with a speed of 1 m/s perpendicular to $\vec{B}$, then $B$ is 1 tesla (T). Also $1\ \text{T} = 10^4$ gauss; the gauss is not an SI unit but is a convenient one.

For scale: the Earth's magnetic field at the surface is about $3.6\times10^{-5}$ T = 0.36 gauss, an MRI magnet 1.5 T to 7 T, and NMR magnets up to 14 T (superconducting electromagnets). High tension power lines produce strong fields; exposure is kept below 0.5 milligauss.

Standard results

• $\vec{F}_m = q(\vec{v}\times\vec{B})$; $F_m = qvB\sin\theta$
• $\vec{F} = q[\vec{E} + (\vec{v}\times\vec{B})]$ (Lorentz force)
• $F_m = 0$ when $v = 0$, or when $\vec{v}$ is parallel or antiparallel to $\vec{B}$
• $F_m$ maximum $= qvB$ when $\vec{v}\perp\vec{B}$
• Magnetic force does zero work; kinetic energy is unchanged
• $1\ \text{T} = 1\ \text{N s C}^{-1}\text{m}^{-1} = 10^4$ gauss

Worked type (textbook Example 10.1)

A negatively charged particle moves through a uniform field in three arrangements. Find $\vec{v}\times\vec{B}$ first by the right-hand rule, then reverse it because the charge is negative. When $\vec{v}$ and $\vec{B}$ are antiparallel the angle is $180^\circ$, $\sin 180^\circ = 0$, and no force acts at all.