Force Between Two Parallel Current Carrying Wires: Magnetic Force, Derivation, Formula, and more

Before getting into the details of magnetic force, a crystal clear understanding of the chronology of events is necessary.

How does a current-carrying wire exert force?

The Biot-Savart law explains that any current-carrying conductor generates a magnetic field around it. The magnetic field depends on several factors, such as the amount of current, the distance from the wire, the length of the wire, and the angle between the current element and the position vector.

Since both wires are carrying current, that means both will produce a magnetic field around themselves. The magnetic interaction between wires allows both wires to exert force on each other.

According to the Lorentz Force, the force on a current-carrying conductor due to a magnetic field can be expressed as; $F = I\,L \times B$$(for\; two\; infinitely\; long\; wires,\; angle\; will\; be\; 90°.)$

The force experienced by a straight wire of length L, carrying current I in a magnetic field B:

$F=\mathrm{<br>}<br>\mathrm{IBL}$

There are two cases when wires are carrying parallel currents to each other.

• Case 1 (Repulsive Force): The direction of flow of current is in the same direction in both wires.
• Case 2 (Attractive Force): The flow of current is in opposite directions to each other in the wires.

We know the formula for the Magnetic field due to a straight current-carrying conductor.

So, the magnitude of the magnetic force on wire 1 due to the magnetic field generated due to wire 2, when the distance between wires is r,

The force per unit length on any of the wires is;

Where;

μ₀ = permitivity of space.

Two key concepts to understand the method to derive the magnetic forcce between two parallel conductors carrying current.

• Right-hand Thumb Rule: The direction of Magnetic Field generated around the wire. 
• Biot-Savart Law: The magnitude Magnetic Field generated around the wire. 

The Ampère’s right-hand grip rule is one of the easiest methods to find the direction of the magnetic field due to a current-carrying wire. What you need is just the direction of the flow of current. 

As per the NCERT Textbooks, "Imagine that you are holding a current-carrying straight conductor in your right hand such that the thumb points towards the direction of current. Then your fingers will wrap around the conductor in the direction of the field lines of the magnetic field." This is known as the right hand thumb rule.

The direction of net magnetic field resultant of both magnetic fields repectively of wire 1 and wire 2 will be the direction of magnetic force.

 

Know more about the right-hand thumb rule through the Wikipedia article on right-hand rule: Right-hand Rule

As per the NCERT textbooks," According to Biot-Savart’s law, the magnitude of the magnetic field dB is proportional to the current I, the element length dl, and inversely proportional to the square of the distance r. Its direction is perpendicular to the plane
containing dl and r."

To calculate the magntic force on a current carrying conductor due to a magnetic field we must know the lorentz force formula. The Lorentz force provides specific formula which gives the magnitude of the actual magnetic force experienced. 

The mathematical expression of lorentz force on a current-carrying conductor;

F=ILBsinθ

Where:

L = Length of the conductor

θ = Angle between current (direction) and magnetic field
.

To derive the formula of the magnetic force between two parallel current-carrying conductors, we need to start with the magnetic field due to a straight wire.

Using Ampère's law for a closed circular loop of radius r around a long straight wire carrying current $I_1$. Since the magnetic field is tangential and constant on the loop, so

$$\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{}$$

$$B(2\pi r)={\mu }_0{I}_1\boxed{{\displaystyle \frac{\mathrm{<br>}\mathrm{<br>}}{}}}$$ $$\oint \mathbf{B}\cdot d\mathbf{l} = B(2\pi r) = \mu_0 I_1 \quad\Rightarrow\quad \boxed{\,B_1(r)=\dfrac{\mu_0 I_1}{2\pi r}\,}$$

Since the Lorentz force on a moving charge $q$

$$\mathbf{f}_{\text{charge}}=q(\mathbf{v}\times\mathbf{B}).$$If this charge is flowing through a wire, considering wire 1 and wire 2, with cross-sectional area $A$, charge carrier density $n$, each carrier charge $q$, and drift velocity $\mathbf{v}_d$.

The total charge flowing through the wire in unit time: $N=nAL$.

So, Net magnetic force on total charge:

$${\mathbf{F}}_{\text{total}}=\mathrm{<br>}\text{nAL}q({\mathbf{v}}_d\times {\mathbf{B}}_1)=\left(nqA{\mathbf{v}}_d\right)L\times {\mathbf{B}}_1\mathrm{.}$$

Since current is charge passing per unit time, So $I_2 = n q A v_d$

$$\boxed{\,\mathbf{F}=I_2\,(\mathbf{L}\times\mathbf{B}_1)\,}.$$

Substitute $\mathbf{B}_1$

$$F = I_2 L B_1 = I_2 L \cdot \frac{\mu_0 I_1}{2\pi r}$$

$$\boxed{\,F=\dfrac{\mu_0 I_1 I_2 L}{2\pi r}\,}$$The force per unit length:

$$\boxed{{\displaystyle \frac{F}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}}}\mathrm{<br>}$$

Here we have provided the NCERT Class 12 notes.