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Force Between Two Parallel Currents (Ampere’s Law)

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Estimated time: 11 minutes
CBSE: Class 12

Introduction

When two long, straight, current-carrying conductors are placed parallel to each other, they exert a magnetic force on each other.

  • If currents flow in the same direction → conductors attract each other.
  • If currents flow in opposite directions → conductors repel each other.
CBSE: Class 12

Definition: Ampere

The ampere is that constant current which, when maintained in each of two infinitely long, straight, parallel conductors of negligible circular cross-section, placed 1 metre apart in a vacuum, produces a force of 2 × 10−7 N per metre of length between them.

CBSE: Class 12

Formula: Force Between Two Parallel Current-Carrying Conductors

\[F=\frac{\mu_0I_1I_2}{2\pi d}\times l\]

Per unit length:

\[\frac{F}{l}=\frac{\mu_0I_1I_2}{2\pi d}\]

Force acts along the line joining the wires

CBSE: Class 12

Derivation of Force Between Two Parallel Currents

Consider two long parallel conductors a and b, separated by a distance d.

  • Conductor a carries current Ia
  • Conductor b carries current Ib

Step 1 - Magnetic Field Due to Conductor a at the Location of Conductor b

Using the result from Biot–Savart law for a long straight wire:

Ba = \[\frac {μ_0I_a}{2πd}\]
  • Direction: Perpendicular to the plane containing the two wires (use right-hand rule).
  • μ0 = 4π × 10−7 T·m/A (permeability of free space).

Step 2 - Force on Conductor b Due to Field Ba

A segment of length L of conductor b carrying current Ib​ in field Ba​ experiences a Lorentz force:

Fba = Ib L Ba

Substituting Ba​:

Fba = Ib L ⋅ \[\frac {μ_0I_a}{2π d}\]
Fba = \[\frac {μ_0I_aI_b}{2π d}\] ⋅ L

Step 3 - Force Per Unit Length

\[\frac{F}{L}=\frac{\mu_0I_aI_b}{2\pi d}\]
 
Symbol Meaning SI Unit
μ0 Permeability of free space T·m/A
Ia, Ib Currents in the two conductors Ampere (A)
d Perpendicular distance between conductors metre (m)
F/L Force per unit length N/m

Step 4 - Newton's Third Law Applied

By symmetry, the magnetic field of conductor b exerts an equal and opposite force on conductor a:

\[F_{ab}=\frac{\mu_0I_aI_b}{2\pi d}\cdot L\]
\[\vec{F}_{ba}=-\vec{F}_{ab}\]

This confirms Newton's Third Law: the forces are equal in magnitude and opposite in direction.

Step 5 - Direction Rule

Current Direction Force Type
Same direction (parallel) Attractive
Opposite direction (antiparallel) Repulsive
CBSE: Class 12

Force Depends on Variables

Variable Increased Effect on F/L Reason
Ia​ doubled F/L doubles Direct proportionality
Ib​ doubled F/L doubles Direct proportionality
d doubled F/L halved Inverse proportionality
Medium changed to paramagnetic F/L increases slightly μ increases
CBSE: Class 12

Example

Given:

  • Horizontal component of Earth's magnetic field: B = 3.0 × 10⁻⁵ T
  • Direction of Earth's field: South → North
  • Current in wire: I = 1 A
  • Find: Force per unit length in two cases

Formula used:

f = \[\frac {F}{l}\] = I B sin⁡ θ

where θ = angle between current direction and magnetic field direction.

Case (a): Current flows East → West

  • Step 1: Find the angle between current (E→W) and Earth's field (S→N).
    → East–West is perpendicular to South–North, so θ = 90°
  • Step 2: Apply formula:
    f = I B sin ⁡90° = 1 × 3 × 10−5 × 1 = 3 × 10−5 N/m
Answer: f = 3 × 10⁻⁵ N/m
This is larger than 2 × 10⁻⁷ N/m (the value used to define the ampere), which is why Earth's field must be cancelled out when standardising the ampere.

Case (b): Current flows South → North

  • Step 1: Find the angle between current (S→N) and Earth's field (S→N).
    → Both points in the same direction, so θ = 0°
  • Step 2: Apply formula:
    f = I B sin ⁡0° = I B × 0 = 0
Answer: f = 0 N/m — No force at all!

Test Yourself

Shaalaa.com | Moving Charge and Magnetism part 29 (Ampere)

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