Force on current-carrying conductors: HSC Physics Module 6

Magnitude:

$F = B I L \sin \theta = 0.12 \times 6.0 \times 0.25 \times \sin 90° = 0.18$ N.

Direction: by the right-hand rule, with fingers pointing in the direction of current flow and curling into the page (direction of $B$), the thumb gives the force direction. If the current flows to the right and the field is into the page, the force on the wire is upward.

Reversing the current reverses the force direction. The magnitude is unchanged: $0.18$ N, now downward.

Markers reward the calculation with units, the right-hand-rule reasoning, and the comment that magnitude stays the same when current is reversed.

Force per unit length:

$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi r} = \frac{4 \pi \times 10^{-7} \times 3.0 \times 5.0}{2 \pi \times 0.080}$
$= \frac{6.0\pi \times 10^{-6}}{0.16\pi} = \frac{6.0 \times 10^{-6}}{0.16}$
$= 3.75 \times 10^{-5}$ N/m.

(Step by step: $\mu_0 / 2 \pi = 2 \times 10^{-7}$, so $F/L = 2 \times 10^{-7} \times 3.0 \times 5.0 / 0.080 = 3.75 \times 10^{-5}$ N/m.)

Currents in the same direction attract: each wire sits in the magnetic field of the other, and by the right-hand rule the magnetic force on a wire in the field of the other points toward the other wire when their currents are parallel.

By Newton's third law, both wires feel the same magnitude of force.

Markers reward correct substitution, the attractive direction with reasoning, and explicit mention of Newton's third law.

The wire is perpendicular to the field, so $\theta = 90^{\circ}$ and $\sin\theta = 1$.

$F = BIL\sin\theta = (0.40)(4.5)(0.60)\sin 90^{\circ} = 1.08\ \text{N} \approx 1.1\ \text{N}$.

Marks: one for the correct formula with values substituted, one for the answer stated to two significant figures with the unit newton.

$\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi r} = \dfrac{(4\pi \times 10^{-7})(4.0)(6.0)}{2\pi (0.050)}$

Using $\mu_0/(2\pi) = 2 \times 10^{-7}\ \text{T m A}^{-1}$:

$\dfrac{F}{L} = \dfrac{(2 \times 10^{-7})(4.0)(6.0)}{0.050} = 9.6 \times 10^{-5}\ \text{N/m}$.

Marks: one for the correct formula (using $\mu_0/2\pi = 2\times10^{-7}$) with values substituted, one for the answer $9.6 \times 10^{-5}$ N/m with the unit.

$F = BIL\sin\theta = (0.80)(3.0)(0.50)\sin 40^{\circ} = (1.2)(0.643) = 0.77\ \text{N}$.

The force is a maximum at $\theta = 90^{\circ}$ (current perpendicular to the field), where $\sin\theta = 1$.

Marks: one for the correct substitution including $\sin 40^{\circ}$, one for $F = 0.77$ N with the unit, one for correctly stating $\theta = 90^{\circ}$ gives the maximum.

(a) Rearrange $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi r}$ for $r$:

$r = \dfrac{\mu_0 I_1 I_2}{2\pi (F/L)} = \dfrac{(2 \times 10^{-7})(10)(15)}{5.0 \times 10^{-5}} = \dfrac{3.0 \times 10^{-5}}{5.0 \times 10^{-5}} = 0.60\ \text{m}$.

(b) The currents flow in the same direction, so the wires attract: each wire lies in the other's magnetic field, and the right-hand rule shows the resulting force on each wire points toward the other wire.

Marks: one for correctly rearranging for $r$, one for substituting with $\mu_0/2\pi = 2\times10^{-7}$, one for $r = 0.60$ m with the unit, one for the correct attract/repel call with right-hand-rule reasoning.

(a) $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi r} = \dfrac{(2 \times 10^{-7})(25)(25)}{0.15} = \dfrac{1.25 \times 10^{-4}}{0.15} = 8.3 \times 10^{-4}\ \text{N/m}$.

(b) $F = (F/L) \times L = (8.3 \times 10^{-4})(6.0) = 5.0 \times 10^{-3}\ \text{N}$.

Marks: one for the correct force-per-length formula with substitution, one for $F/L = 8.3 \times 10^{-4}$ N/m, one for multiplying by the $6.0$ m length, one for $F = 5.0 \times 10^{-3}$ N with the unit.

(a) The graph is a straight line through the origin, so $F/L$ is directly proportional to $I_2$ ($F/L \propto I_2$), consistent with $F/L = \mu_0 I_1 I_2/(2\pi r)$ with $\mu_0$, $I_1$ and $r$ held constant.

(b) Gradient $= \dfrac{\Delta (F/L)}{\Delta I_2} = \dfrac{(1.6 \times 10^{-4} - 4.0 \times 10^{-5})\ \text{N/m}}{(8.0 - 2.0)\ \text{A}} = \dfrac{1.2 \times 10^{-4}}{6.0} = 2.0 \times 10^{-5}\ \text{N m}^{-1}\text{A}^{-1}$.

(c) The gradient of $F/L$ against $I_2$ equals $\dfrac{\mu_0 I_1}{2\pi r}$:

$\dfrac{\mu_0 I_1}{2\pi r} = \dfrac{(2 \times 10^{-7})(5.0)}{0.050} = \dfrac{1.0 \times 10^{-6}}{0.050} = 2.0 \times 10^{-5}\ \text{N m}^{-1}\text{A}^{-1}$,

which matches the gradient found from the graph in part (b).

Marks: one for identifying direct proportionality (line through the origin), one for a correctly calculated gradient with units, one for correctly identifying the gradient as $\mu_0 I_1/(2\pi r)$, one for substituting $I_1$ and $r$, one for showing the two values agree.

Band-6 plan. (1) State the parallel-wire force law and what it depends on. (2) Explain, using the right-hand rule, why parallel currents attract and antiparallel currents repel. (3) Show algebraically that reversing one current's direction leaves $F/L$ unchanged in magnitude, only its sense (toward/away) changes. (4) Conclude the claim is correct and state the physical reason (the force depends on $I_1 I_2$ as a product of magnitudes).

Model answer. The force per unit length between two long parallel wires is $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi r}$, where $I_1$ and $I_2$ are the magnitudes of the two currents. This expression depends only on the magnitudes of the currents, their separation $r$, and the constant $\mu_0$ - reversing the direction of a current does not change its magnitude, so the size of $F/L$ is unaffected by which way either current flows.

What does change is the direction of the force. Each wire sits in the magnetic field produced by the other. Using the right-hand rule, when the two currents flow in the same direction, the field of one wire at the location of the other, combined with $F = I L \times B$, gives a force on each wire directed toward the other wire - the wires attract. If one current is reversed, the field direction at the second wire is unchanged, but the direction of $I_2$ in $F = I_2 L B \sin\theta$ (vector sense) flips, so the force on that wire reverses to point away from the first wire - the wires now repel.

Therefore the student's claim is correct: reversing the current in one wire reverses the direction of the magnetic force between the two wires (attraction becomes repulsion, or vice versa) but leaves the magnitude $F/L = \mu_0 I_1 I_2/(2\pi r)$ unchanged, because that magnitude depends on the product of the current magnitudes, not their directions.

Marker's note: the top band explicitly separates magnitude (unchanged, because $F/L$ depends on $|I_1||I_2|$) from direction (reversed, argued from the right-hand rule applied to each wire in the other's field), and states the attract-to-repel consequence. A response that only asserts "same direction attracts, opposite repels" without the magnitude argument caps in the middle band.

Band-6 plan. (1) State the motor-effect force law for one wire in an external field. (2) Derive $B$ produced by a long straight wire and substitute into the motor-effect law to obtain $F/L = \mu_0 I_1 I_2/(2\pi r)$ - showing it is not a new law. (3) Discuss the historical significance (pre-2019 SI definition of the ampere). (4) Reach an evaluative judgement.

Model answer. The motor effect states that a wire of length $L$ carrying current $I$ in an external uniform magnetic field $B$ experiences a force $F = BIL\sin\theta$, perpendicular to both the current and the field. This law describes the force on any current-carrying conductor placed in a magnetic field, regardless of the field's source.

A long straight current-carrying wire itself produces a magnetic field. Wire 1, carrying current $I_1$, produces a field at a perpendicular distance $r$ of magnitude $B_1 = \dfrac{\mu_0 I_1}{2\pi r}$. If a second wire carrying current $I_2$ lies in this field, perpendicular to it, the motor-effect law applies directly to wire 2: $\dfrac{F}{L} = B_1 I_2 \sin 90^{\circ} = \dfrac{\mu_0 I_1 I_2}{2\pi r}$. By Newton's third law, wire 1 experiences an equal and opposite force per unit length from the field of wire 2. This derivation shows the "parallel-wire force" is not an independent law - it is the motor-effect force applied to a wire sitting in the field generated by another current-carrying wire.

This relationship carries real historical significance: before the 2019 SI redefinition, the ampere itself was defined using this exact result. One ampere was the constant current that, flowing in each of two infinite parallel wires one metre apart in a vacuum, would produce a force per unit length of exactly $2 \times 10^{-7}$ N/m between them - a value fixed by defining $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$ exactly. The whole SI system of electrical units was therefore anchored to a mechanical force measurement derivable from the single-wire motor-effect law.

On balance, the claim is well supported: the parallel-wire formula is mathematically and physically a direct application of $F = BIL\sin\theta$ to the field generated by a second current, not a separate phenomenon, and this unifying derivation is precisely what made the parallel-wire force useful enough to define the base SI unit of current.

Marker's note: the top band completes the actual derivation (substituting $B_1 = \mu_0 I_1/(2\pi r)$ into $F = BIL\sin\theta$), not just quoting both formulas side by side, and connects the result to the historical ampere definition with the exact $2\times10^{-7}$ N/m value. A response that states both laws without the substitution step caps in the middle band.