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# Force between Two Parallel Current Carrying Conductors

Moving charges produce an electric field and the rate of flow of charge is known as current. This is the basic concept in Electrostatics. The magnetic effect of electric current is the other important phenomenon related to moving electric charges. Magnetism is generated due to the flow of current. Magnetic fields exert force on the moving charges and at the same time on other magnets, all of which have moving charges. When the charges are stationary, their magnetic field doesn’t affect the magnet but when charges move, they produce magnetic fields that exert force on other magnets.

Magnetism is produced by the movement of charges around a conductor. Magnetism is often a property exhibited by magnets and caused by moving charges that force objects to be pulled or pushed away.

## Force in a Magnetic Field

The movement of charges generates a magnetic field and the magnetic force exerted in that field is referred to as the force produced by the magnetic field. The fundamental property of matter that allows it to produce and experience electrical and magnetic effects is called charge. The magnetic field of a magnet is a specific area in space where the magnet exerts its magnetic effect. Assume that there is a point charge q that is present in the magnetic field B (r) and the electric field E (r) and that it is traveling at a velocity of v while being placed at r at a particular time t. The force exerted by both of them on an electric charge q can be expressed as,

F = q [E(r) + v × B(r)] = FElectric + Fmagnetic

This formula was stated by H.A. Lorentz for the force due to the electric field, based on the extensive experiments of Ampere and others. It is also called the Lorentz force.

## Force between Two Parallel Current Carrying Conductor

Magnetic field is generated by a current-carrying conductor. Another current-carrying conductor experiences force as a result of the external magnetic field. Therefore, we can say that any two current-carrying conductors will exert a magnetic force on one another when they are put close to each other.

The forces between two parallel currents are of two types:

• Attractive: When current is flowing in the same direction in both wires then attractive force is exerted.
• Repulsive: When current is flowing in the opposite direction in both wires then repulsive force is exerted. Consider two parallel current-carrying wires, separated by a distance ‘d’, such that one of the wires is carrying current I1 and the other is carrying I2. From previous studies, we can say that wire 2 experiences the same magnetic field at every point along its length due to wire 1. Using the Right-Hand Thumb rule we can determine the direction of magnetic force.

The magnitude of the field due to the first conductor can be calculated using Ampere’s Circuital Law by,

Ba = μ0I1 / 2πd

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

F21 = I2LB1 = (μ0I1I2 / 2πd) 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,

F12 = -F21

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

Fba = μ0IaIb / (2πd)

where,
d is the distance between two conducor
Ia is the current in wire 1
Ib is the current in wire 2

## Force between Two Parallel Current Carrying Sheets

Consider two parallel Sheets carrying currents producing a uniform magnetic field of induction B between the planes. Outside this space, there is no magnetic field. The magnetic force acting per unit area of each plane is

F = B2 / 2μ0

The magnetic field due to one of the sheets is (1/2) B. The force on the sheet is given by,

F = (1/2)B × i × Length × Breadth

F = (B2 / 2μ0) per unit area.

The above-mentioned formula is similar to F = BIl on a straight wire.

Also, Check

## Solved Examples on Current-Carrying Wire

Example 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 fab = fba = f = 1.5 × 10-4 N

distance (d) = 4.8m

The force per unit length on wires is given as,

fab = fba = f = μ0IaIb / 2πd           —(1)

when the current in both wires is doubled,

I’a = 2Ia

I’b = 2Ib

Distance between the wires is halved,

d’ = d/2

equation (1) can be written as,

f’ab = f’ba = f’ = μ0I’aI’b / 2πd’

f’ = 2 × (μ0×2Ia×2Ib / 2πd)

f’ = 8 × (μ0×Ia×Ib / 2πd)

f’ = 8f

f’ = 8 × 1.5 × 10-4 N

f’ = 12 × 10-4 N

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

Solution:

Current in both the wires Ia = Ib = 6A

distance (d) = 3m

The force per unit length is given as,

fab = fba = f = μ0IaIb / 2πd           —(1)

from equation (1),

f = (μ0×6×6) / (2π×3)

f = (6μ0/π) × (4/4)

f = 24μ0 / 4π                —(2)

we know,

μ0/4π = 10-7 T – m/A        —(3)

from equation (2) and (3),

f = 24 × 10-7

Example 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 and find the magnetic force between the two wires.

Solution:

Given that

Current in the first wire Ia = 8 A

Current in the second wire Ib = 11 A

Distance between two wires d = 15 m

F/L = μ0 × Ia × Ib /(2πd)

= 4π x 10-7 × 8 × 11/(2π × 15)

= 176 × 10-7/15

=11.733 × 10-7 N

Therefore, the magnetic force between two wires is 11.733 × 10-7 N.

Example 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:

Current in wire A, I1 =5 A

Current in wire B, I2 =3 A

Force exerted on 10cm section of wire A,

B = μ0 × 2 × I1× I2 / (4πd)

= 4π x 10-7 × 2 × 5 × 3/(4π × 6 × 10-2)

= 5 × 10-5 T

The Attractive force is normal from A towards B, because current direction is same.

Example 5: Two wires carry currents of  50 A and  70 A respectively and they repel each other with a force of  0.25 N/m. The distance between them will be

Solution:

Current in wire A, I1 = 50 A

Current in wire B, I2 = 70 A

Magnetic Force, F = 0.25 N/m

F = μ0 × 2 × I1× I2 / (4πd)

0.25 = 4π x 10-7 × 2 × 50 × 70/(4π × d)

0.25 × d = 7 × 10-4

d = 2.8 × 10-3 m

## FAQs on Current-Carrying Wire

### Question 1: Define Magnetic field.

The magnetic field or magnetic induction by that material or by that current is the area or space around the current-carrying wire/moving electric charge or surrounding the magnetic object in which force of magnetism can be felt by other magnetic materials.

### Question 2: Determine the force between two current-carrying wires.

Ampere’s law states that, if 1 ampere of current flows through each of two parallel conductors of infinite length, separated by 1 meter in empty space free from other magnetic fields, causes a force of exactly 2 × 10−7 N/m on each conductor.

### Question 3: What is the relationship between the currents in the two wires?

The two wires carrying current in the same direction attract each other, and they repel each other if the currents flows in opposite direction.

### Question 4: Which force is acting between two parallel current-carrying conductors?

An external magnetic field exerts a force on a current-carrying conductor and the Lorentz force formula governs this principle. Thus, we can say that any two current carrying conductors when placed near each other, will exert a magnetic force on each other.