Investigate quantitatively and analyse the interaction between current-carrying conductors and uniform magnetic fields F/l = I B sin theta, including parallel current-carrying wires F/l = mu_0 I_1 I_2 / (2 pi r)

What this dot point is asking

NESA wants you to apply $F = BIL \sin \theta$ to a straight wire carrying current $I$ of length $L$ in a uniform field $B$, derive the parallel-wire force $F/L = \mu_0 I_1 I_2 / (2 \pi r)$ as a special case, work out direction with the right-hand rule, and connect the parallel-wire result to the historical definition of the ampere.

The answer

Force on a straight wire in a uniform field

A wire of length $L$ carrying current $I$ in a uniform magnetic field $\vec{B}$ experiences a force:

where $\theta$ is the angle between the current direction and the field. Per unit length:

This force comes from the magnetic force on each moving charge in the wire: $\vec{F} = q \vec{v} \times \vec{B}$ summed over $N$ charges gives $\vec{F} = I \vec{L} \times \vec{B}$ in vector form.

Direction is given by the right-hand rule:

1. Point the fingers of the right hand in the direction of the conventional current.
2. Curl them toward $\vec{B}$.
3. The thumb gives the force direction.

Equivalently: flat right hand, fingers along $\vec{B}$, thumb along $I$, palm pushes the force out.

Special angles

• $\theta = 90°$: maximum force, $F = BIL$.
• $\theta = 0°$: zero force (current parallel to field).
• $\theta = 180°$: zero force (current antiparallel to field).

Force between two long parallel wires

Two long, straight, parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $r$ exert magnetic forces on each other. The magnetic field at wire 2 due to wire 1 (at distance $r$) is:

This field is perpendicular to wire 2, so the force per unit length on wire 2 is:

By Newton's third law, wire 1 feels the same magnitude of force per unit length.

The constant $\mu_0 = 4 \pi \times 10^{-7}$ T m/A is the permeability of free space. With this value, $\mu_0 / (2 \pi) = 2 \times 10^{-7}$ T m/A exactly, which simplifies a lot of arithmetic.

Attraction and repulsion

• Same direction currents. Each wire sits in the magnetic field of the other; the right-hand rule shows the force on each wire points toward the other wire. The wires attract.
• Opposite direction currents. The forces reverse. The wires repel.

This is sometimes summarised as "parallel currents attract, antiparallel currents repel," the reverse of the rule for electric charges.

Historical definition of the ampere

The pre-2019 SI definition of the ampere used parallel wires. One ampere was defined as the current in each of two infinite, parallel wires 1 m apart in vacuum that would produce a force per unit length of:

This defined $\mu_0 = 4 \pi \times 10^{-7}$ T m/A exactly. Since the 2019 SI redefinition the ampere is defined via the fixed value of the electronic charge $e$, and $\mu_0$ is measured rather than defined, but the value is essentially unchanged for HSC work.

Worked example: rail gun (qualitative)

A conducting bar of length $L = 0.30$ m slides along two rails carrying a current $I = 200$ A. It sits in a field $B = 0.50$ T perpendicular to both the bar and the rails. The force on the bar is:

$F = BIL = 0.50 \times 200 \times 0.30 = 30$ N.

This force accelerates the bar along the rails. Rail guns scale this idea up to thousands of amperes and tesla-class fields to launch projectiles.

Try it: Lorentz force calculator for the force on moving charges, the same physics that gives $F = BIL\sin\theta$ when summed over a current.

Examples in context

Example 1. Force on a TransGrid 330 kV transmission conductor in Earth's magnetic field. A north-south section of the TransGrid Liddell-to-Bayswater 330 kV line carries $I = 1500 \text{ A}$ in a horizontal span of $L = 400 \text{ m}$ at right angles to Earth's field $B = 5.0 \times 10^{-5} \text{ T}$. Force on the span is $F = B I L \sin 90^{\circ} = 5.0 \times 10^{-5} \times 1500 \times 400 = 30 \text{ N}$, directed vertically (down for one half-cycle, up for the other in AC). At 50 Hz this gives a sub-audible mechanical hum but is far too small ($30 \text{ N}$ on a $\sim 1000 \text{ kg}$ span) to affect catenary sag.

Example 2. Force between parallel busbars in a Snowy 2.0 generator hall. Two parallel busbars carry $I_1 = I_2 = 12{,}000 \text{ A}$ in the same direction and are separated by $r = 0.30 \text{ m}$. Force per unit length is $F/L = \mu_0 I_1 I_2 / (2 \pi r) = 4 \pi \times 10^{-7} \times (1.2 \times 10^4)^2 / (2 \pi \times 0.30) = 96 \text{ N/m}$, attractive. Over a $10 \text{ m}$ run, the busbars try to pull together with $960 \text{ N}$, equivalent to a $98 \text{ kg}$ mass hanging between them. This is why Snowy Hydro generators use heavy ceramic-insulated supports every $1.5 \text{ m}$ along the busbar runs.

Try this

Q1. Write the formula for the force on a current-carrying wire in a magnetic field and define each symbol. [2 marks]

• Cue. $F = B I L \sin\theta$; $B$ field strength (T), $I$ current (A), $L$ length in field (m), $\theta$ angle between $I$ and $B$.

Q2. A straight wire of length $0.40 \text{ m}$ carries $5.0 \text{ A}$ at $30^{\circ}$ to a $0.25 \text{ T}$ field. Calculate the force on the wire. [3 marks]

• Cue. $F = 0.25 \times 5.0 \times 0.40 \times \sin 30^{\circ} = 0.25 \text{ N}$.

Q3. Two parallel wires $0.10 \text{ m}$ apart each carry $20 \text{ A}$ in the same direction. (a) Calculate the force per unit length between them. (b) State whether the force is attractive or repulsive. (c) Use this result to explain how the ampere was historically defined. [3+1+2 marks]

• Cue. (a) $F/L = \mu_0 I^2 / (2\pi r) = 4\pi \times 10^{-7} \times 400 / (2\pi \times 0.10) = 8.0 \times 10^{-4} \text{ N/m}$. (b) Attractive. (c) Old SI: $1 \text{ A}$ produces $2 \times 10^{-7} \text{ N/m}$ between $1 \text{ m}$-apart wires.