Question

Two long straight parallel conductors A and B carrying steady currents $I_a$ and $I_b$ in the same direction are separated by a distance d. Deduce the expressions for the force acting on length L of conductor B due to conductor A and show it in figure. Write the expression for the force acting on length L of conductor A due to conductor B and show that it follows Newton's third law.

Hint:

A useful mnemonic for parallel wires: "Like currents Attract, Unlike currents Repel." This behaves totally oppositely to standard electrostatic charges where "Like charges repel".

Answer 1

Concept:
• A long, straight current-carrying wire universally generates a cylindrical magnetic field circulating around it, described exactly by Ampere's Circuital Law.
• When a second parallel wire carrying its own current is placed firmly within this generated magnetic field, it aggressively experiences a magnetic Lorentz force.
• The definitive direction of this mutual force is strictly governed by Fleming's Left-Hand Rule or the right-hand cross product rule.

Step 1: Calculate the magnetic field created by Conductor A

\includegraphics[width=0.5\linewidth]{22OR.png}
Let conductor A strongly carry a steady current $I_a$ flowing directly upwards.
Conductor B is placed parallel to A at a fixed perpendicular distance $d$.
According to Ampere's Circuital Law, the magnitude of the magnetic field $\vec{B}_a$ produced uniquely by conductor A precisely at the location of conductor B is:
\[ B_a = \frac{\mu_0 I_a}{2\pi d} \]
Using the Right-Hand Grip Rule, if current $I_a$ flows upwards, the magnetic field lines aggressively circulate counter-clockwise. At the exact position of wire B (located to the right of A), this magnetic field vector $\vec{B}_a$ points strictly perpendicular and into the plane of the paper/page.

Step 2: Deduce the force acting on length L of Conductor B
Conductor B rigidly carries a steady current $I_b$, also flowing directly upwards in the same direction.
A specific length $L$ of this conductor firmly sitting inside the external magnetic field $\vec{B}_a$ will continuously experience a magnetic force $\vec{F}_{BA}$.
The foundational formula for this magnetic force is rigorously given by:
\[ \vec{F}_{BA} = I_b (\vec{L} \times \vec{B}_a) \]
Since the straight length vector $\vec{L}$ (pointing upwards) and the magnetic field vector $\vec{B}_a$ (pointing strictly inwards) are perfectly mutually perpendicular ($\theta = 90^\circ$), the scalar magnitude is:
\[ F_{BA} = I_b L B_a \sin(90^\circ) = I_b L B_a \]
Substitute the previously derived expression for $B_a$ into the force equation:
\[ F_{BA} = I_b L \left( \frac{\mu_0 I_a}{2\pi d} \right) \]
\[ F_{BA} = \frac{\mu_0 I_a I_b L}{2\pi d} \]
Applying Fleming's Left-Hand Rule (Current up, Field strictly inwards), the resulting force $F_{BA}$ on wire B points squarely towards the left, pulling directly towards conductor A. Thus, it is an attractive force.

Step 3: Expression for force on Conductor A and Newton's Third Law
By applying the exact same rigorous logic in reverse, conductor B physically creates a magnetic field $B_b = \frac{\mu_0 I_b}{2\pi d}$ pointing out of the page at the specific location of conductor A.
The force $F_{AB}$ experienced by a length $L$ of conductor A strictly due to this field is:
\[ F_{AB} = I_a L B_b = I_a L \left( \frac{\mu_0 I_b}{2\pi d} \right) = \frac{\mu_0 I_a I_b L}{2\pi d} \]
Applying Fleming's Left-Hand Rule again (Current up, Field strictly outwards), the resulting force $F_{AB}$ on wire A points squarely towards the right, pulling directly towards conductor B.
Mathematically comparing the two derived vector forces, their absolute magnitudes are strictly identical:
\[ |F_{BA}| = |F_{AB}| = \frac{\mu_0 I_a I_b L}{2\pi d} \]
However, their physical directions are perfectly exactly opposite to each other. Conductor A pulls B to the left, while Conductor B pulls A to the right.
In strict vector notation:
\[ \vec{F}_{BA} = -\vec{F}_{AB} \]
This explicitly and conclusively proves that the mutual magnetic forces actively existing between the two parallel conductors rigorously obey Newton's Third Law of Motion (action and equal, opposite reaction).