ScholarQuill logoScholarQuillUniversity Notes
  • Notes
  • Past Papers
  • Blogs
  • Todo
Login
ScholarQuill logoScholarQuillUniversity Notes
Login
NotesPast PapersBlogsTodo
More
SubjectsDiscussionCGPA CalculatorGPA CalculatorStudent PortalCourse Outline
About
About usPrivacy PolicyReportContact
Notes
Past Papers
Blogs
Todo
Analytics
    Current Subject
    🧩
    Applied Physics
    GE-169
    Progress0 / 45 topics
    Topics
    1. Electric Force and Its Applications2. Conservation of Charge3. Charge Quantization4. Electric Fields Due to Point Charge and Lines of Force5. Electric Fields: Ring of Charge and Disk of Charge6. A Point Charge in an Electric Field7. Dipole in an Electric Field8. Flux of a Vector Field9. Flux of an Electric Field10. Gauss’ Law and Its Applications11. Spherically Symmetric Charge Distribution12. Charge Isolated Conductor13. Electric Potential Energy14. Electric Potentials and Related Problems15. Calculating Potential from the Field16. Potential Due to Point and Continuous Charge Distribution17. Potential Due to a Dipole18. Equipotential Surfaces19. Calculating the Field from the Potential20. Electric Current and Current Density21. Resistance, Resistivity, and Conductivity22. Ohm's Law and Its Applications23. The Hall Effect24. Magnetic Force on a Current25. The Biot-Savart Law26. Line of Magnetic Field (B)27. Two Parallel Conductors28. Ampere's Law29. Solenoids and Toroids30. Faraday's Experiments and Law of Induction31. Lenz's Law32. Motional EMF33. Induced Electric Fields34. The Basic Equations of Electromagnetism35. Induced Magnetic Fields36. The Displacement Current37. Reflection and Refraction of Light Waves38. Total Internal Reflection39. Two Source Interference40. Double-Slit Interference and Related Problems41. Interference from Thin Films42. Diffraction and Wave Theory43. Single-Slit Diffraction and Related Problems44. Polarization of Electromagnetic Waves45. Polarizing Sheets and Related Problems
    GE-169›Two Parallel Conductors
    Applied PhysicsTopic 27 of 45

    Two Parallel Conductors

    9 minread
    1,542words
    Intermediatelevel

    Two Parallel Conductors: Concept and Applications

    When two conductors are placed parallel to each other and carry electric currents, they experience forces due to the magnetic fields they generate. This setup plays a significant role in understanding electromagnetic interactions, such as the force between current-carrying wires, the concept of magnetic fields due to currents, and applications in electrical engineering like the operation of motors and transformers.

    1. Magnetic Field Due to a Current-Carrying Conductor

    First, let's recall the magnetic field created by a long, straight, current-carrying conductor. The magnetic field at a distance rrr from a long straight conductor carrying a current III is given by Ampère's Law (or Biot-Savart Law) as:

    B=μ0I2πrB = \frac{\mu_0 I}{2 \pi r}B=2πrμ0​I​

    Where:

    • BBB is the magnetic field at a distance rrr from the conductor,
    • μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}μ0​=4π×10−7T\cdotpm/A is the permeability of free space,
    • III is the current in the conductor,
    • rrr is the radial distance from the conductor to the point where the field is measured.

    This magnetic field is circular around the conductor and its direction is given by the right-hand rule: If you point the thumb of your right hand in the direction of the current, your fingers curl in the direction of the magnetic field.

    2. Magnetic Force Between Two Parallel Conductors

    When two parallel conductors carry currents, they each produce magnetic fields that interact with the other conductor. This interaction results in a force between the conductors. The nature of this force (attractive or repulsive) depends on the direction of the currents in the wires.

    Force per Unit Length on a Conductor Due to the Other Conductor

    The force per unit length F/LF/LF/L on one current-carrying conductor due to the magnetic field of the second conductor is derived from the Lorentz force law, which states that the force on a current-carrying wire is the product of the current, the length of the wire, and the magnetic field. For two conductors carrying currents I1I_1I1​ and I2I_2I2​, and separated by a distance ddd, the force per unit length is:

    FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}LF​=2πdμ0​I1​I2​​

    Where:

    • FFF is the force between the conductors,
    • LLL is the length of the conductors under consideration,
    • I1I_1I1​ and I2I_2I2​ are the currents in the two conductors,
    • ddd is the distance between the conductors,
    • μ0\mu_0μ0​ is the permeability of free space.

    Direction of the Force

    The direction of the force is given by the right-hand rule for magnetic forces:

    • If the currents in the two conductors are in the same direction, the force between them is attractive.
    • If the currents in the two conductors are in opposite directions, the force between them is repulsive.

    This is because the magnetic fields generated by each current interact in such a way that like currents attract and opposite currents repel.

    3. Applications of Force Between Parallel Conductors

    The interaction between current-carrying conductors has several practical applications in physics and engineering:

    1. Ampere's Law and the Definition of the Ampere

    The force between two parallel conductors forms the basis for the definition of the ampere. By measuring the force per unit length between two conductors carrying known currents, we can define the unit of current (the ampere). The ampere is defined as the current that, when flowing through two parallel conductors 1 meter apart, produces a force of 2×10−7 N/m2 \times 10^{-7} \, \text{N/m}2×10−7N/m between them.

    2. Transmission Lines

    In electrical power transmission, conductors are used to carry electric current over long distances. The force between parallel transmission lines can affect the mechanical structure of the lines and their stability. Understanding these forces is crucial for designing structures that can handle mechanical stresses due to magnetic interactions between the conductors.

    3. Electromagnetic Relays and Motors

    Many electromagnetic devices, such as motors and relays, depend on the force between parallel current-carrying conductors. In electric motors, two sets of conductors (the rotor and stator) interact with each other through magnetic fields, resulting in mechanical motion. Similarly, in relays, the interaction between conductors causes the relay to switch, activating or deactivating circuits.

    4. Magnetic Levitation (Maglev)

    In maglev trains, the force between current-carrying conductors is used to lift and propel the train. By using electromagnetic forces, the train can be levitated and moved along a track without physical contact, reducing friction and enabling high-speed travel.

    4. Example Problems

    Example 1: Force Between Two Parallel Conductors

    Problem: Two long, parallel conductors are placed 0.5 meters apart. The current in the first conductor is 10 A10 \, \text{A}10A and the current in the second conductor is 20 A20 \, \text{A}20A. What is the force per unit length between the two conductors?

    Solution:

    We use the formula for the force per unit length between two parallel conductors:

    FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}LF​=2πdμ0​I1​I2​​

    Substitute the given values:

    • μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}μ0​=4π×10−7T\cdotpm/A,
    • I1=10 AI_1 = 10 \, \text{A}I1​=10A,
    • I2=20 AI_2 = 20 \, \text{A}I2​=20A,
    • d=0.5 md = 0.5 \, \text{m}d=0.5m.
    FL=(4π×10−7)(10)(20)2π(0.5)\frac{F}{L} = \frac{(4\pi \times 10^{-7})(10)(20)}{2\pi(0.5)}LF​=2π(0.5)(4π×10−7)(10)(20)​

    Simplifying:

    FL=8×10−61=8×10−6 N/m\frac{F}{L} = \frac{8 \times 10^{-6}}{1} = 8 \times 10^{-6} \, \text{N/m}LF​=18×10−6​=8×10−6N/m

    So, the force per unit length between the two conductors is 8×10−6 N/m8 \times 10^{-6} \, \text{N/m}8×10−6N/m.

    Example 2: Direction of Force Between Two Current-Carrying Conductors

    Problem: Two long, parallel conductors carry currents of 5 A5 \, \text{A}5A each. If the currents are flowing in opposite directions, determine the direction of the force between them.

    Solution:

    • Since the currents are in opposite directions, the magnetic fields around each conductor will interact in such a way that the conductors will repel each other.
    • The force between the conductors will be repulsive.

    To determine the direction of the force using the right-hand rule:

    • For the first conductor, point your right thumb in the direction of the current and curl your fingers around the conductor to visualize the magnetic field.
    • For the second conductor, apply the same rule to determine the direction of the magnetic field at the location of the first conductor.

    The resulting force between the conductors will be repulsive because the currents are in opposite directions.


    Summary

    1. Magnetic field due to a current-carrying conductor: The magnetic field around a conductor carrying current III is given by B=μ0I2πrB = \frac{\mu_0 I}{2 \pi r}B=2πrμ0​I​.
    2. Force between two parallel conductors: The force per unit length between two conductors carrying currents I1I_1I1​ and I2I_2I2​ separated by a distance ddd is FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}LF​=2πdμ0​I1​I2​​.
    3. Direction of the force: The force is attractive if the currents are in the same direction and repulsive if the currents are in opposite directions.
    4. Applications: This principle is used in defining the ampere, designing transmission lines, and understanding the operation of electromagnets, motors, and other devices based on magnetic interactions.

    Understanding the force between parallel conductors is fundamental in both theoretical electromagnetism and practical engineering applications involving current-carrying wires and electromagnetic devices.

    Previous topic 26
    Line of Magnetic Field (B)
    Next topic 28
    Ampere's Law

    Past Papers

    Open this section to load past papers

    Click on Show Past Papers to see past papers.
    On This Page
      Reading Stats
      Est. reading time9 min
      Word count1,542
      Code examples0
      DifficultyIntermediate