Moving charges generate 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.

The movement of charges generates magnetism around a conductor. Generally, magnetism is a property shown by magnets and produced by moving charges, which results in objects being attracted or pushed away.

## Magnetic Field

When a charge moves, it creates a magnetic field. And the force created in a magnetic field is called Magnetic Force. A charge is a basic property associated with the matter due to which it produces and experiences electrical and magnetic effects. A particular region in space around the magnet where the magnet has its magnetic effect is called the magnetic field of the magnet. Assume that there is a point charge q (moving with a velocity v and, located at r at given time t) in presence of both the electric field E (r) and the magnetic field B (r). The force on an electric charge q due to both of them can be written as,

F = q [E(r) + v × B(r)] ≡ E_{Electric}+ F_{magnetic}

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

A magnetic field is created around a conductor due to the current flowing through it. An external magnetic field exerts a force on a current-carrying conductor. Thus, we can say that any two current-carrying conductors when placed near each other, will exert a magnetic force on each other.

There are two types of Force between two parallel currents:

**Attractive:**If the currents have the same direction or Currents are flowing in the same direction.**Repulsive:**If the currents are in the opposite direction or they are flowing in the opposite direction.

Consider two parallel current-carrying conductors, separated by a distance ‘d’, such that one of the conductors is carrying a current I_{1} and the other is carrying I_{2}. From previous studies, we can say that conductor 2 experiences the same magnetic field at every point along its length due to conductor 1. The direction of magnetic force can be found using the right-hand thumb rule.

From Ampere’s circuital law, the magnitude of the field due to the first conductor can be given by,

**B _{a} = μ_{0}I_{1} / 2πd**

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

**F _{21} = I_{2}LB_{1} = (μ_{0}I_{1}I_{2} / 2πd) L**

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

**F _{12} = -F_{21}**

Also, the currents flowing in the same direction make the conductors attract each other and that showing in the opposite direction makes the conductors repel each other. The magnitude of the force acting per unit length can be given as,

F_{ba}= μ_{0}I_{a}I_{b}/ 2πdwhere,

dis the distance between two conducor,I_{a}_{ }is the current in a conductor,I_{b}_{ }is the current in b conductor,

## Solved Problems on Force between Two Parallel Current Carrying Conductors

**Problem 1: Two long parallel wires separated by 0.1 m carry currents of 1A and 2A respectively in opposite directions. A third current-carrying wire parallel to both of them is placed in the same plane such that it feels no net magnetic force. It is placed at a distance of **

**Solution: **

We know that the magnetic field due to long straight wire,

B = μ

_{0}I / 2πxTherefore, B

_{1}= B_{2}(μ_{d}/ 2πx) = μ_{d2}/ 2π(0.1 + x)Here, i

_{1}= 1Ai

_{2}= 2Aμ

_{0}×1 / 2πx = μ_{0}×2 / 2π(0.1 + x)2x = (0.1 + x)

x = 0.1 m

**Problem 2: The force per unit length is 10 ^{-3} N on the two current-carrying wires of equal length that are separated by a distance of 2 m and placed parallel to each other. If the current in both the wires is doubled and the distance between the wires is halved, then the force per unit length on the wire will be? **

**Solution: **

Force per unit length on both wires f

_{ab}= f_{ba}= f = 10^{-3}Ndistance (d) = 2m

The force per unit length on wires is given as,

f—(1)_{ab}= f_{ba}= f = μ_{0}I_{a}I_{b}/ 2πdwhen the current in both wires is doubled,

I’

_{a}= 2I_{a}I’

_{b}= 2I_{b}Distance between the wires is halved,

d’ = d/2

equation (1) can be written as,

f’

_{ab}= f’_{ba}= f’ = μ_{0}I’_{a}I’_{b}/ 2πd’f’ = 2 × (μ

_{0}×2I_{a}×2I_{b}/ 2πd)f’ = 8 × (μ

_{0}×I_{a}×I_{b}/ 2πd)f’ = 8f

f’ = 8 × 10-3 N

**Problem 3: Two very long wires are placed parallel to each other and separated by a distance of 1m apart. If the current in both the wires is 1A, then the force per unit length on both wires will be: **

**Solution: **

Current in both the wires I

_{a}= I_{b}= 1Adistance (d) = 1m

The force per unit length is given as,

f—(1)_{ab}= f_{ba}= f = μ_{0}I_{a}I_{b}/ 2πdfrom equation (1),

f = (μ

_{0}×1×1) / (2π×1)f = μ

_{0}/2π × 2/2f = 2μ

_{0}/4π —(2)we know,

μ

_{0}/4π = 10^{-7}T – m/A —(3)from equation (2) and (3),

f = 2 × 10

^{-7 }N

**Problem 4: The length of two wires is 0.5 m and the distance between the wires is 1m. If 1 A current is passed in the wires in the same direction, the force per unit length between the wires is: **

**Solution:**

Length of two wires (L1 and L2) = 0.5m

Distance between the wires (d) = 1m

Current flowing in each wire (I

_{1}and I_{2}) = 1AMagnetic field produced on wire 1 by wire 2 is,

B

_{21}= μ_{0}I_{1}/ 2πr = 4π×10^{-7}×1 / 2π×1 = 2 × 10^{-7}TMagnetic field produced by wire 2 on wire 1 is,

B

_{12}= μ_{0}I_{2}/ 2πr = 4π×10-7×1 / 2π×1 = 2 × 10^{-7}TUsing Fleming’s left-hand rule,

Force (F

_{1}) and (F_{2}) is acting on wire 1 and 2 respectively, B_{12}is directed to the right side and B_{21}is directed to the left side.Using, F = ILB

F

_{1}= I_{1}L_{1}B_{12}= 1 × 0.5 × (2 x 10^{-7}) = 1 × 10^{-7}NF

_{2}= I_{2}L_{2}B_{21}= 1 × 0.5 × (2 x 10^{-7}) = 1 × 10^{-7}NSince, F1 = F2 and they are directed opposite to one another, the net force is zero.

Hence, the force per unit length between the wires is also zero.

**Problem 5: Wire P carrying current 6 A upward and wire Q is 1m apart from it. If μ _{0} = 4π×10^{-7} wb A^{-1} m^{-1} and there is a repulsive force between wire P and Q 1.2×10^{-5} N.m^{-1}. Determine the magnitude and direction of electric current on wire Q. **

**Solution: **

Current (I

_{p}) = 6 Aμ

_{0}= 4π × 10^{-7}wb A^{-1}m^{-1}Repulsive force (F) = 1.2×10

^{-5}N m^{-1}L = 1 m

Electric current on wire is given by,

F = (μ

_{0}/2π) (I_{p}I_{q}/L)1.2 × 10

^{-5}= (4π×10^{-7}/2π) (6×I_{q}/1)1.2 × 10

^{-5}= (2×10^{-7}) (6×I_{q})1.2 × 10

^{-5}= (12×10^{-7}) (I_{q})1.2 = (12×10

^{-2}) (I_{q})1.2 = 0.12(I

_{q})I

_{q}= 1.2 / 0.12I

_{q}= 10 A

**Problem 6: Wire A and B are 1m apart. Wire P carrying current 1A. If μ _{0} = 4π×10^{-7} Wb.A^{-1}.m^{-1} and there is an attractive force on each other 10^{-7} N.m^{-1}, determine the magnitude and direction of electric current on wire Q.**

**Solution: **

Current (I

_{p}) = 1 Aμ

_{0}= 4π × 10^{-7}wbA^{-1}m^{-1}Attractive force (F) = 10

^{-7}Nm^{-1}Length between the wire (L) = 1 m

Electric current on wire is given by,

F = (μ

_{0}/2π) (I_{p}I_{q}/L)10

^{-7}= (4π×10^{-7}/ 2π) (1×I_{q}/ 1)10

^{-7}= (2×10^{-7}) (I_{q})1 = 2(I

_{q})I

_{q}= 1/2I

_{q}= 0.5 A

## FAQs on Force between Two Parallel Current Carrying Conductors

**Question 1: State Ampere’s Circuit Law. **

**Answer:**

Ampere’s circuital law states that “the line integral of the magnetic field surrounding closed-loop equals to the number of times the algebraic sum of currents passing through the loop.

**Question 2: Explain the nature of parallel and anti-parallel currents.**

**Answer: **

Due to a current carrying conductor their exists a magnetic field around it. And an external magnetic field exerts a force on a current-carrying conductor. Therefore we can say that when two current-carrying conductors are placed near each other they will exert magnetic forces on each other. When the current flows in the same direction in the two parallel wires then both wires attract each other and if the current flows in the opposite direction in the two parallel wires then both wires repel each other.

**Question 3: Define Magnetic field **

**Answer: **

The space or region around the current carrying wire/moving electric charge or around the magnetic material in which force of magnetism can be experienced by other magnetic material is called magnetic field or magnetic induction by that material or by that current.