Ampere force and current interactions

The Ampere force is the force exerted on a current-carrying conductor in a magnetic field. For a straight segment of length $l$ in a uniform field with induction $\vec B$, it is $\vec F = I\,\vec l \times \vec B$, and its magnitude is $F = B I l \sin\alpha$, where $\alpha$ is the angle between the current direction and the vector $\vec B$. The direction of the force is found by the left-hand rule: place the left hand so that the lines of $\vec B$ enter the palm and the four fingers point along the current; then the extended thumb indicates the direction of $\vec F$. For two long parallel conductors with currents $I_1$ and $I_2$ separated by a distance $d$, the force per unit length is $F/l = \mu_0 I_1 I_2/(2\pi d)$: currents in the same direction attract, while opposite currents repel. If the field is produced by several sources, first find the resultant induction $\vec B$ by the superposition principle, and then calculate the Ampere force.

A straight conductor of length $l = 0{,}20$ m carrying current $I = 5{,}0$ A is placed in a uniform magnetic field with induction $B = 0{,}40$ T. The conductor is perpendicular to the vector $\vec B$. Find the magnitude of the Ampere force and state its direction qualitatively.

Two long parallel conductors are separated by a distance $d = 0{,}10$ m. The first carries current $I_1 = 10$ A, and the second carries current $I_2 = 5{,}0$ A in the same direction. Find the force per unit length on each conductor and state whether they attract or repel.

Two long parallel conductors are separated by a distance $d = 0{,}20$ m. They carry currents $I_1 = 6{,}0$ A and $I_2 = 4{,}0$ A in opposite directions. Exactly halfway between them, a third long parallel conductor is placed with current $I_3 = 3{,}0$ A directed the same way as $I_1$. Find the resultant force per unit length on the third conductor and its direction.

Use a safe computer simulation (for example, PhET or an interactive model of two parallel conductors). Set up two long parallel conductors and vary the currents $I_1$, $I_2$, and the distance $d$. For each set, measure the force per unit length and verify the dependence $F/l \propto I_1 I_2/d$. Then reverse the direction of one current and confirm that attraction changes to repulsion. Record the results in a table and plot $F/l$ against $1/d$ for constant currents.

Show the direction of the Ampere force and the left-hand rule, as well as the attraction and repulsion of parallel currents.

Diagram: a straight conductor with current $I$ in a uniform field $\vec B$ (field lines and force vector $\vec F$); next to it, two parallel conductors with currents $I_1$ and $I_2$, with forces $\vec F_1$ and $\vec F_2$ shown for the same-direction and opposite-direction cases.

In a phone, a tiny DC motor spins an eccentric mass. Its rotor coil sits in a radial magnetic field $B = 0.25\ \text{T}$. The two active sides of the coil each have length $L = 0.012\ \text{m}$ and carry $I = 0.08\ \text{A}$. Each side feels a force $F = B I L = 0.25 \cdot 0.08 \cdot 0.012 = 2.4\times10^{-4}\ \text{N}$; the pair forms a torque about the axis. At $12000\ \text{rpm}$ ($200\ \text{Hz}$) this torque produces the buzzing vibration.

This uses the Ampère force law $F = B I L \sin\theta$ for a straight conductor segment in a uniform field; here $\theta = 90^\circ$ for each active side, and the two opposite forces form a couple (torque) — the same principle as in any electric motor.

In a radio's loudspeaker, a voice coil of total wire length $L = 6.0\ \text{m}$ sits in a radial magnetic field $B = 0.85\ \text{T}$. An audio signal drives $I = 1.2\ \text{A}$ through the coil. The total Ampère force is $F = B I L = 0.85 \cdot 1.2 \cdot 6.0 = 6.1\ \text{N}$ (all segments contribute with $\sin\theta=1$), pushing the cone forward and back at the audio frequency. For a $100\ \text{Hz}$ bass note the cone moves about $2\ \text{mm}$ peak-to-peak.

The voice coil is a many-turn loop; each short segment obeys $d\mathbf{F} = I\, d\mathbf{l} \times \mathbf{B}$. Because the field is radial, every segment's force is axial, so the vector sum is $F = B I L$, the integral form of the Ampère force.

At a substation, two parallel aluminium busbars are $d = 0.12\ \text{m}$ apart and carry $I_1 = I_2 = 4000\ \text{A}$ in the same direction during peak load. Each metre of each bar feels a force per unit length

$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{(4\pi\times10^{-7}) (4000)(4000)}{2\pi (0.12)} \approx 26.7\ \text{N/m}.$$

The bars are attracted toward each other. If the currents were opposite (as in a short circuit), the same magnitude would repel them, and the supports must withstand that force.

This is the interaction of currents: each current creates a magnetic field $B = \mu_0 I/(2\pi d)$ at the other wire, and the second wire feels $F = B I L$. The superposition of fields is used to isolate the field of one wire at the location of the other.