Slide 1

  • Topic: More Applications of Ampere’s Law - Magnetic field outside solenoid
  • Introduction to Ampere’s Law and its applications
  • Recap of the magnetic field inside a solenoid
  • Overview of the topic coverage in this lecture
  • Learning objectives for the lecture

Slide 2

  • Revisiting Ampere’s Law:
    • States the relationship between the magnetic field and the current enclosed by a closed loop
    • Formula: ∮B ⋅ dl = μ₀I
  • Magnetic field outside a long straight conductor:
    • Formula: B = (μ₀I) / (2πr)
  • Comparing the magnetic fields inside and outside a straight conductor
  • Examples of calculating the magnetic field outside a conductor

Slide 3

  • Magnetic field outside a solenoid:
    • Formula: B = μ₀nI
  • Understanding the parameters in the formula: n and I
  • Comparison of the magnetic field inside and outside a solenoid
  • Examples of calculating the magnetic field outside a solenoid

Slide 4

  • Magnetic field outside a toroid:
    • Formula: B = (μ₀nI) / (2πR)
  • Understanding the parameters in the formula: n, I, and R
  • Explaining the concept of a toroid and its magnetic field
  • Examples of calculating the magnetic field outside a toroid

Slide 5

  • Magnetic field outside a current-carrying loop:
    • Formula: B = (μ₀I) / (2R)
  • Understanding the parameters in the formula: I and R
  • Explanation of the orientation of the magnetic field outside a loop
  • Examples of calculating the magnetic field outside a current-carrying loop

Slide 6

  • Magnetic field due to a long straight conductor and a current-carrying loop
  • Superposition principle for magnetic fields
  • Using Ampere’s Law to calculate the total magnetic field
  • Examples showcasing the application of superposition principle

Slide 7

  • Magnetic field outside a long solenoid with iron core
  • Introduction to magnetic materials and their properties
  • Explanation of how magnetic materials affect the magnetic field
  • Understanding the role of iron core in increasing the magnetic field
  • Examples illustrating the magnetic field outside a long solenoid with iron core

Slide 8

  • Magnetic field outside a current sheet
  • Derivation of the formula: B = (μ₀I) / (2h)
  • Understanding the parameters in the formula: I and h
  • Explaining the concept of a current sheet and its magnetic field
  • Examples of calculating the magnetic field outside a current sheet

Slide 9

  • Magnetic field outside a current-carrying circular loop
  • Derivation of the formula: B = (μ₀IR²) / (2z(z² + R²)^(3/2))
  • Understanding the parameters in the formula: I, R, and z
  • Explaining the orientation and magnitude of the magnetic field
  • Examples of calculating the magnetic field outside a current-carrying circular loop

Slide 10

  • Magnetic field of a finite straight conductor
  • Derivation of the formula: B = (μ₀I) / (4π) * (sina + sinb) / (r)
  • Understanding the parameters in the formula: I, a, b, and r
  • Explaining the concepts of finite conductor and magnetic field
  • Examples of calculating the magnetic field of a finite straight conductor

Slide 11

  • Summary of previous examples and formulas covered
  • Magnetic field outside a solenoid: B = μ₀nI
    • n: number of turns per unit length
    • I: current through the solenoid
  • Key points to remember:
    • The magnetic field is directly proportional to the current and the number of turns per unit length.
    • The magnetic field outside the solenoid is very weak compared to the field inside.
  • Example 1: A solenoid has 500 turns per meter and carries a current of 2 A. Calculate the magnetic field outside the solenoid.

Slide 12

  • Magnetic field outside a toroid: B = (μ₀nI) / (2πR)
    • R: radius of the toroid
  • Key points to remember:
    • The magnetic field is directly proportional to the current and the number of turns per unit length.
    • The magnetic field inside the toroid is zero.
  • Example 1: A toroid has 500 turns and a radius of 0.2 m. If it carries a current of 4 A, calculate the magnetic field outside the toroid.

Slide 13

  • Magnetic field outside a current-carrying loop: B = (μ₀I) / (2R)
    • R: distance from the center of the loop to the point where the field is calculated
  • Key points to remember:
    • The magnetic field is inversely proportional to the distance from the center of the loop.
    • The magnetic field outside the loop is weaker than inside.
  • Example 1: A circular loop of radius 0.1 m carries a current of 5 A. Calculate the magnetic field at a distance of 0.2 m from the center of the loop.

Slide 14

  • Superposition principle for magnetic fields: The total magnetic field at a point due to multiple current-carrying conductors is the vector sum of the magnetic fields produced by each conductor individually.
  • The magnetic field due to each conductor can be calculated using the formulas discussed earlier.
  • Key points to remember:
    • Vector addition should be used when calculating the total magnetic field.
    • The direction of each individual magnetic field should be considered while adding them.
  • Example 1: Two long straight conductors are placed parallel to each other. One carries a current of 3 A in the upward direction, and the other carries a current of 5 A in the downward direction. Calculate the total magnetic field at a point equidistant from both conductors.

Slide 15

  • Magnetic field outside a long solenoid with an iron core
  • Magnetic materials exhibit higher magnetic permeability than vacuum or air, amplifying the magnetic field when placed inside a solenoid.
  • Key points to remember:
    • Using an iron core increases the magnetic field strength outside the solenoid.
    • The core concentrates the magnetic field lines, enhancing it.
  • Example 1: A long solenoid without a core produces a magnetic field of 0.5 T. If an iron core is inserted into the solenoid, calculate the new magnetic field outside the solenoid.

Slide 16

  • Magnetic field outside a current sheet: B = (μ₀I) / (2h)
    • h: perpendicular distance from the center of the sheet to the point where the field is calculated
  • Key points to remember:
    • The magnetic field is inversely proportional to the distance from the current sheet.
    • The field lines are parallel to the sheet and form concentric circles.
  • Example 1: A current sheet carries a current of 10 A. Find the magnetic field at a distance of 0.1 m from the sheet.

Slide 17

  • Magnetic field outside a current-carrying circular loop: B = (μ₀IR²) / (2z(z² + R²)^(3/2))
    • z: distance from the center of the loop to the point where the field is calculated
  • Key points to remember:
    • The magnetic field is inversely proportional to the distance from the center of the loop.
    • The field lines are perpendicular to the plane of the loop and form concentric circles.
  • Example 1: A circular loop of radius 0.05 m carries a current of 2 A. Calculate the magnetic field at a distance of 0.1 m from the center of the loop.

Slide 18

  • Magnetic field of a finite straight conductor: B = (μ₀I) / (4π) * (sina + sinb) / (r)
    • a, b: angles subtended by the conductor at the point where the field is calculated
  • Key points to remember:
    • The magnetic field is inversely proportional to the distance from the conductor.
    • The angles a and b represent the limits of the conductor length contributing to the magnetic field at the point.
  • Example 1: A finite straight conductor of length 0.05 m carries a current of 3 A. Calculate the magnetic field at a point 0.1 m away from one end of the conductor.

Slide 19

  • Summary of different scenarios covered
  • Recap of the formulas for the magnetic field outside various current-carrying conductors
  • Understanding the relationships between the parameters in each formula
  • Key points to remember:
    • Magnetic fields outside conductors are weaker than inside.
    • The distance from the conductor influences the strength of the field.
    • Superposition principle applies when multiple conductors are present.
  • Example 1: A straight conductor, a solenoid, and a circular loop, each carrying a current of 4 A, are placed in close proximity. Calculate the total magnetic field at a point equidistant from all three conductors.

Slide 20

  • Questions from the audience
  • Encourage the students to ask questions related to the lecture content
  • Provide clarification or elaboration wherever necessary
  • Summary of key takeaways from the lecture
  • Reminder of any upcoming assignments or assessments
  • Thank the students for their participation and attention during the lecture

Slide 21

  • Summary of Ampere’s Law and its applications
  • Recap of the magnetic field inside a solenoid
  • Overview of the topic coverage in this lecture
  • Learning objectives for the lecture

Slide 22

  • Magnetic field outside a long straight conductor:
    • Formula: B = (μ₀I) / (2πr)
    • Example: A long straight conductor carries a current of 5 A. Calculate the magnetic field at a distance of 0.1 m from the conductor.
  • Comparing the magnetic fields inside and outside a straight conductor

Slide 23

  • Magnetic field outside a solenoid:
    • Formula: B = μ₀nI
    • Example: A solenoid has 200 turns per meter and carries a current of 3 A. Calculate the magnetic field outside the solenoid.
  • Examples of calculating the magnetic field outside a solenoid

Slide 24

  • Magnetic field outside a toroid:
    • Formula: B = (μ₀nI) / (2πR)
    • Example: A toroid has 300 turns and a radius of 0.1 m. If it carries a current of 2 A, calculate the magnetic field outside the toroid.
  • Examples of calculating the magnetic field outside a toroid

Slide 25

  • Magnetic field outside a current-carrying loop:
    • Formula: B = (μ₀I) / (2R)
    • Example: A circular loop of radius 0.05 m carries a current of 4 A. Calculate the magnetic field at a distance of 0.2 m from the center of the loop.
  • Examples of calculating the magnetic field outside a current-carrying loop

Slide 26

  • Superposition principle for magnetic fields
  • Adding magnetic fields due to multiple current-carrying conductors
  • Example: Two long straight conductors are placed parallel to each other. One carries a current of 6 A in the upward direction, and the other carries a current of 3 A in the downward direction. Calculate the total magnetic field at a point equidistant from both conductors.

Slide 27

  • Magnetic field outside a long solenoid with iron core
  • The role of magnetic materials in enhancing the magnetic field
  • Example: A long solenoid without a core produces a magnetic field of 0.3 T. If an iron core is inserted into the solenoid, calculate the new magnetic field outside the solenoid.

Slide 28

  • Magnetic field outside a current sheet
    • Formula: B = (μ₀I) / (2h)
    • Example: A current sheet carries a current of 8 A. Find the magnetic field at a distance of 0.2 m from the sheet.
  • Examples of calculating the magnetic field outside a current sheet

Slide 29

  • Magnetic field outside a current-carrying circular loop
    • Formula: B = (μ₀IR²) / (2z(z² + R²)^(3/2))
    • Example: A circular loop of radius 0.08 m carries a current of 3 A. Calculate the magnetic field at a distance of 0.1 m from the center of the loop.
  • Examples of calculating the magnetic field outside a current-carrying circular loop

Slide 30

  • Magnetic field of a finite straight conductor
    • Formula: B = (μ₀I) / (4π) * (sina + sinb) / (r)
    • Example: A finite straight conductor of length 0.08 m carries a current of 2 A. Calculate the magnetic field at a point 0.15 m away from one end of the conductor.
  • Examples of calculating the magnetic field of a finite straight conductor