Force on Electric Wire
When a current-carrying conductor is placed in a magnetic field, a force acts on it. Experimentally, one finds a proportionality between the force and the length of the conductor, as well as the current and the magnetic flux density. The force is largest when the current flows exactly perpendicular to the field lines. If the current flows parallel to the field lines, no force occurs. For arbitrary directions, only the projection of the conductor perpendicular to the field lines needs to be considered. The magnitude of the force is therefore given by:
$$
F = I l B \sin \alpha
$$
where $\alpha$ is the angle between the current direction and the magnetic field lines. To generalize further, we consider not only the length $l$ of the conductor but also its direction. In this case, the formula can be written as the magnitude of the cross product:
$$
\boxed{\vec{F} = I \vec{l} \times \vec{B}}
$$
This formulation also provides the direction of the resulting force.
Example:
A current-carrying wire is placed in a magnetic field with a flux density of 50 mT. The portion of the wire inside the field is 10 cm long, and the current is 2 A. What is the force on the wire if it makes an angle of $60^\circ$ with the field lines?
Solution: $8.6 \,\text{mN}$
Since a current-carrying conductor also generates a magnetic field, it exerts a force on another conductor. This interaction has been used since 1948 to define the unit ampere: it is the current that produces a force of $2 \cdot 10^{-7}$ N per meter of length between two infinitely long parallel conductors one meter apart.
Force on Single Charge
Replacing the current with its definition
$$
I = \frac{\mathrm{d} q}{\mathrm{d} t}
$$
and combining with the length element, one arrives at the general formulation of the Lorentz force:
$$
\boxed{\vec{F_L} = q \vec{v} \times \vec{B}}
$$
This gives the force on a charge $q$ moving with velocity $\vec{v}$ through a magnetic field. For a single electron with elementary charge $e$:
$$
F_L = e \vec{v} \times \vec{B}
$$
The Lorentz force always acts perpendicular to both the velocity and the magnetic field lines. Its direction can be determined using the right-hand rule (for positive charges) or left-hand rule (for negative charges), also known as the three-finger rule:
- Index finger → direction of the magnetic field
- Thumb → velocity of the charge
- Middle finger → direction of the force
Motion of Charged Particle
If a charge moves perpendicular to the field lines, the Lorentz force is maximal and acts as a radial force, forcing the particle onto a circular trajectory. By equating Lorentz force and centripetal force,
$$
F_L = F_Z
$$
we obtain:
$$
q v B = \frac{m v^2}{r}
$$
Solving for the radius $r$ gives:
$$
\boxed{R = \frac{m v}{q B}}
$$
Example:
A proton moves at 1% of the speed of light, perpendicular to the field lines of a magnetic field with flux density 100 mT. What is the radius of the resulting circular path?
Solution: $31.4 \,\text{cm}$
If the particle’s velocity has both perpendicular and parallel components relative to the field, the perpendicular component causes circular motion, while the parallel component remains unaffected. The resulting path is a helix (spiral) with a constant radius.
Experiment: Beam Tube Experiment
The Lorentz force can be impressively demonstrated using a beam tube experiment. A glass sphere is filled with a low-density gas and placed between a pair of coils generating a homogeneous magnetic field. Electrons are first accelerated by an electric field and then enter the apparatus with constant velocity, oriented perpendicular to the field lines. Inside the sphere, the electrons move along a circular path. Collisions with the gas make the trajectory visible as a faint glow. In a darkened room, a glowing ring can clearly be observed.
The principle of the Lorentz force is applied in mass spectrometry, where charged particles are deflected in a magnetic field. Their radius of curvature depends on their mass-to-charge ratio, allowing different ions to be separated and analyzed.