Force between Two Parallel Current Carrying Conductors (original) (raw)

Last Updated : 4 Jun, 2026

Moving charges produce an electric field, and the rate at which charge flows is called "electric current," a fundamental concept in electromagnetism. A current-carrying conductor generates a magnetic field around it, and when another current-carrying conductor is placed nearby, it experiences a force due to this magnetic field. Hence, any two current-carrying conductors exert a magnetic force on each other when brought close.

The forces between two parallel currents are of two types:

Force between Two Parallel Current Carrying Conductor

Magnetic Force Between Two Parallel Current-Carrying Conductors

Consider two parallel current-carrying wires separated by a distance d, where one wire carries current I1 and the other carries current I2. From previous studies, wire 2 experiences a uniform magnetic field at every point along its length due to wire 1. The direction of this magnetic field and the resulting force can be determined using the Right-Hand Thumb Rule.

The magnitude of the magnetic field produced by the first conductor at the position of the second conductor is given by Ampere’s Circuital Law as

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

The force on a segment of length L of wire 2 due to wire 1 can be given as,

F_{21} = I_2 L B_1 = \left(\frac{\mu_0 I_1 I_2}{2\pi d}\right) L

Similarly, we can calculate the force exerted by wire 2 on wire 1. We see that wire 1 experiences the same force due to wire 2, but the direction is opposite. Thus,

F_{12} = -F_{21}

Also, the currents flowing in the same direction make the wires attract each other, and those flowing in the opposite direction make the wires repel each other. We can find the magnitude of the force acting per unit length by the formula,

\frac{F}{L} = \frac{\mu_0 I_a I_b}{2\pi d}

Where:

Force between Two Parallel Current Carrying Sheets

Consider two parallel current-carrying sheets producing a uniform magnetic field of induction B in the region between them, while the magnetic field outside this region is zero. The magnetic force acting per unit area on each sheet is given by:

F = \frac{B^2}{2\mu_0}

The magnetic field due to one of the sheets is \frac{B}{2}​. The force acting on a sheet can be expressed as

F = \left(\frac{B}{2}\right) \times i \times \text{Length} \times \text{Breadth}

This expression is analogous to the force on a straight current-carrying conductor given by \text{F = B I L}

Solved Problems

**Question 1: Two current-carrying wires of equal length are parallel to one another and spaced 4.8 m apart, producing a force of 1.5 10⁻⁴ N per unit length. What will be the force per unit length on the wire if the current in both wires is doubled and the distance between the wires is halved?

**Solution: Force per unit length on both wires,

f_{ab} = f_{ba} = f = 1.5 \times 10^{-4} \, \text{N}

Distance,

d = 4.8 \, \text{m}

The force per unit length is given by,

f = \frac{\mu_0 I_a I_b}{2\pi d} \quad ...(1)

When the current in both wires is doubled,

I'_a = 2I_a, \quad I'_b = 2I_b

Distance between the wires is halved,

d' = \frac{d}{2}

Using equation (1),

f' = \frac{\mu_0 I'_a I'_b}{2\pi d'}

f' = \frac{\mu_0 (2I_a)(2I_b)}{2\pi (d/2)}

f' = \frac{4\mu_0 I_a I_b}{2\pi} \times \frac{2}{d}

f' = 8 \times \frac{\mu_0 I_a I_b}{2\pi d}

f' = 8f

f' = 8 \times 1.5 \times 10^{-4}

f' = 12 \times 10^{-4} \, \text{N}

**Question 2: Two very long wires are placed parallel to each other and separated by a distance 3m apart. If the current in both wires is 6A, then the force per unit length on both wires will be:

**Solution: Current in both wires,

I_a = I_b = 6A

Distance,

d = 3\,m

The force per unit length is given by,

f_{ab} = f_{ba} = f = \frac{\mu_0 I_a I_b}{2\pi d} \quad ...(1)

From equation (1),

f = \frac{\mu_0 \times 6 \times 6}{2\pi \times 3}

f = \frac{36\mu_0}{6\pi}

f = \frac{6\mu_0}{\pi}

\frac{\mu_0}{4\pi} = 10^{-7} \quad .....(3)

From equation (2) and (3),

f = 24 \times 10^{-7} \, \text{N}

**Question 3: If 8 A of current flows in the first wire, 11 A of current flows in the second wire. The distance between two wires is 15 m; find the magnetic force between the two wires.

**Solution: Force per unit length is given by,

\frac{F}{L} = \frac{\mu_0 I_a I_b}{2\pi d}

= \frac{4\pi \times 10^{-7} \times 8 \times 11}{2\pi \times 15}

= \frac{4 \times 8 \times 11 \times 10^{-7}}{2 \times 15}

= \frac{352 \times 10^{-7}}{30}

= 11.733 \times 10^{-7} \, \text{N/m}

**Question 4: Two long and parallel straight wires A and B carrying currents of 5 A and 3 A in the same direction are separated by a distance of 6 cm. Estimate the force on a 12 cm section of wire A.

**Solution: Force between two parallel wires is given by,

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

Substituting values,

F = \frac{4\pi \times 10^{-7} \times 5 \times 3}{2\pi \times 6 \times 10^{-2}} \times 12 \times 10^{-2}

F = \frac{60 \times 10^{-7}}{12 \times 10^{-2}} \times 12 \times 10^{-2}

F = 60 \times 10^{-7}

F = 6 \times 10^{-6} \, \text{N}

Since currents are in the same direction, the force is attractive (wire A is attracted towards wire B).

Unsolved Problems

**Question 1: Two long parallel wires carry currents of 4 A and 6 A in the same direction and are separated by a distance of 5 cm. Find the force per unit length between them.

**Question 2: Two parallel conductors carry currents of 10 A and 15 A in opposite directions and are placed 8 cm apart. Calculate the force per unit length between them and state its nature.

**Question 3: Two wires carry equal currents and are placed 10 cm apart. If the force per unit length between them is 2 × 10-4 N/m, find the current flowing in each wire.

**Question 4: Two long parallel wires separated by 20 cm carry currents in the ratio 2:3. If the force per unit length between them is 3 × 10-5 N/m, find the currents in the wires.

**Question 5: Two parallel wires carrying currents of 5 A and 8 A are separated by a distance of 12 cm. Find the force acting on a 15 cm length of one of the wires.

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