(a) Deduce the expression for the force between two long parallel current carrying conductors. **OR** (b) Writ

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Q. (a) Deduce the expression for the force between two long parallel current carrying conductors. OR

(b) Write down Maxwell equations in integral form.

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Concept understanding — Force Between Parallel Wires

Force Between Parallel Current-Carrying Wires

Imagine two long, straight wires placed side by side, each carrying an electric current. You already know that a current-carrying wire creates a magnetic field around it. And you know that a wire placed in a magnetic field experiences a magnetic force. So here, each wire sits inside the magnetic field created by the other wire. That is the whole story — each wire feels a force because of the other wire's magnetic field.

The direction of that force — attraction or repulsion — depends on whether the currents flow in the same direction or opposite directions.

The Intuition

Take two wires with currents in the same direction. Use the right-hand thumb rule: for wire 1, the magnetic field lines circle around it. At the location of wire 2, that field points in a particular direction. Now apply the right-hand rule for force on a current-carrying wire (Fleming's left-hand rule works too): the current in wire 2, crossed with the field from wire 1, gives a force toward wire 1. The same reasoning from wire 2's perspective gives a force on wire 1 toward wire 2. So they attract.

If the currents are opposite, the field directions reverse, and the forces point away from each other — they repel.

Tip

A quick memory aid: Same direction → Attract; Opposite direction → Repel. This is the opposite of what you might guess from electric charges, where like charges repel. Don't mix them up.

The Precise Statement

For two long, straight, parallel wires separated by a distance $d$ , carrying steady currents $I_1$ and $I_2$ , the magnitude of the force per unit length on either wire is:

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}

where $\mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2$ is the permeability of free space.

The force is attractive if the currents are in the same direction, repulsive if they are opposite.

Where Does This Formula Come From?

Wire 1 produces a magnetic field at the location of wire 2. The magnitude of that field is:

B_1 = \frac{\mu_0 I_1}{2 \pi d}

This field is perpendicular to wire 2. The magnetic force on a length $L$ of wire 2 carrying current $I_2$ in a perpendicular field $B_1$ is:

F = I_2 L B_1

Substitute $B_1$ :

F = I_2 L \cdot \frac{\mu_0 I_1}{2 \pi d}

Divide both sides by $L$ to get force per unit length:

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}

That is the entire derivation — two simple steps: field from one wire, then force on the other.

Important

This formula assumes the wires are infinitely long (or at least very long compared to $d$ ) and thin. It gives the force per unit length, which is constant along the wires.

The Definition of the Ampere

This effect is so fundamental that it defines the SI unit of current. One ampere is defined as the constant current which, when flowing through two infinitely long, straight, parallel wires of negligible cross-section placed one metre apart in vacuum, produces a force of exactly $2 \times 10^{-7}$ newtons per metre of length between them. …

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(a) Using $B=\mu_0I_1/(2\pi d)$ from wire 1 and $F=BI_2l$ on wire 2 gives a force per unit length $\mu_0I_1I_2/(2\pi d)$ ; (b) Maxwell's four equations relate E and B fields to their sources in integral form. Both alternatives answered below.

(a) Force between two long parallel current-carrying conductors

Consider two long, straight, parallel conductors separated by distance $d$ , carrying currents $I_1$ and $I_2$ .

The magnetic field produced by wire 1 (carrying $I_1$ ), at the location of wire 2 (distance $d$ away):

$B_1=\dfrac{\mu_0I_1}{2\pi d}$

This field is (nearly) uniform along the length of wire 2, and perpendicular to it. The force experienced by a length $l$ of wire 2 (carrying current $I_2$ ) in this field:

$F=B_1I_2l=\dfrac{\mu_0I_1I_2l}{2\pi d}$

So the force per unit length between the two wires is

$f=\dfrac{F}{l}=\dfrac{\mu_0I_1I_2}{2\pi d}$

By Newton's third law, wire 1 experiences an equal and opposite force due to wire 2's field. If the currents are in the same direction, the wires attract; if in opposite directions, they repel. (This force law is used to define the SI unit of current, the ampere.)

(b) Maxwell's equations in integral form

Gauss's law for electricity (relates electric flux to enclosed charge): $\oint \vec E\cdot d\vec A=\dfrac{Q_{enc}}{\varepsilon_0}$
Gauss's law for magnetism (no magnetic monopoles — magnetic flux through any closed surface is zero): $\oint \vec B\cdot d\vec A=0$
Faraday's law of electromagnetic induction (a changing magnetic flux induces an emf/electric field): $\oint \vec E\cdot d\vec l=-\dfrac{d\Phi_B}{dt}$ …

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Q.(a) Deduce the expression for the force between two long parallel current carrying conductors. OR

(a) Force between two long parallel current-carrying conductors

(b) Maxwell's equations in integral form

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