Induksi Elektromagnetik • Part 1: Fluks Magnetik

Jendela Sains
27 Oct 202008:40

Summary

TLDRIn this video, the concept of magnetic flux is introduced as part of the study of electromagnetic induction. The video explains magnetic flux in comparison to electric flux, detailing the formula and its components, including magnetic field strength (B), area (A), and the angle between the magnetic field and the normal to the surface. Two example problems are solved: one calculating the magnetic flux through a circular area with a given magnetic field and angle, and another calculating the flux through a solenoid. This video offers valuable insights for students learning physics at the high school level.

Takeaways

  • 😀 Magnetic flux is similar to electric flux, representing the amount of magnetic field lines passing through a surface.
  • 😀 The formula for magnetic flux is: Φ = B * A * cos(α), where B is the magnetic field strength, A is the area, and α is the angle between the magnetic field and the normal to the surface.
  • 😀 Magnetic field lines are represented by blue lines, and the surface area is represented by green squares.
  • 😀 The angle α refers to the angle between the magnetic field and the normal of the surface, not between the field and the surface itself.
  • 😀 In a given example, the magnetic field is 0.1 Tesla, and the area is a circle with a radius of 1.4 meters, resulting in a magnetic flux of 0.308 Weber.
  • 😀 The formula for flux calculation uses the area of a circle: A = π * r^2, where r is the radius.
  • 😀 If the magnetic field is at an angle of 30° to the surface, the angle with the normal is 60° (90° - 30°).
  • 😀 For calculating flux in a solenoid, the formula B = μ₀ * (N * I) / L is used, where μ₀ is the permeability of free space, N is the number of turns, I is the current, and L is the length of the solenoid.
  • 😀 The flux through the solenoid's center is calculated using the formula Φ = B * A * cos(α). For the given problem, the flux is 5 × 10^-2 Weber.
  • 😀 The magnetic field strength inside the solenoid is found to be 50 Tesla, calculated using the known values for permeability, number of turns, current, and length.

Q & A

  • What is the definition of magnetic flux?

    -Magnetic flux is the total number of magnetic field lines (or magnetic field strength) passing through a given area. It is similar to electric flux, where the flux represents the total number of electric field lines passing through a surface.

  • How is magnetic flux calculated?

    -Magnetic flux is calculated using the formula: Φ_B = B * A * cos(α), where Φ_B is the magnetic flux, B is the magnetic field strength in Tesla, A is the area in square meters, and α is the angle between the magnetic field lines and the normal to the surface.

  • What is the role of the normal in the formula for magnetic flux?

    -In the formula for magnetic flux, the normal refers to a line perpendicular to the surface area being considered. The angle α is measured between the magnetic field and the normal, not the surface itself.

  • Why is the angle α measured between the magnetic field and the normal?

    -The angle α is measured between the magnetic field and the normal to the surface because the flux depends on the component of the magnetic field that is perpendicular to the surface. The normal line provides a consistent reference for this angle.

  • What is the formula for magnetic flux when the magnetic field is perpendicular to the surface?

    -When the magnetic field is perpendicular to the surface, α = 0°, and the formula simplifies to Φ_B = B * A, as cos(0°) = 1.

  • In the first example, what values were provided for the magnetic field strength and area?

    -In the first example, the magnetic field strength (B) was 0.1 Tesla, and the area of the circular surface was calculated using the radius of 1.4 meters, which resulted in an area of 6.16 m².

  • How was the area of the circular surface calculated in the first example?

    -The area of the circular surface was calculated using the formula A = π * r², where r = 1.4 meters. The value of the area turned out to be 6.16 m².

  • What was the magnetic flux value calculated in the first example?

    -The magnetic flux in the first example was calculated to be 0.3 Weber, using the values of magnetic field strength, area, and the angle of 60°.

  • What is the formula for magnetic flux in a solenoid, and how was it applied in the second example?

    -The formula for magnetic flux in a solenoid is Φ_B = B * A * cos(α). In the second example, the magnetic field at the center of the solenoid was calculated using the formula B = μ₀ * (N / L) * I, where μ₀ is the permeability of free space, N is the number of turns, L is the length, and I is the current. This value was then used in the flux formula.

  • How did the second example calculate the magnetic field in the solenoid?

    -In the second example, the magnetic field in the solenoid was calculated using the formula B = μ₀ * (N / L) * I, where μ₀ = 4π × 10⁻⁷ T·m/A, N = 200 turns, L = 4 cm (converted to meters), and I = 10 amperes. The result was a magnetic field strength of 50 Tesla.

Outlines

plate

This section is available to paid users only. Please upgrade to access this part.

Upgrade Now

Mindmap

plate

This section is available to paid users only. Please upgrade to access this part.

Upgrade Now

Keywords

plate

This section is available to paid users only. Please upgrade to access this part.

Upgrade Now

Highlights

plate

This section is available to paid users only. Please upgrade to access this part.

Upgrade Now

Transcripts

plate

This section is available to paid users only. Please upgrade to access this part.

Upgrade Now
Rate This

5.0 / 5 (0 votes)

Related Tags
Electromagnetic InductionMagnetic FluxPhysics EducationHigh School PhysicsScience TutorialsElectromagnetismMagnetic FieldPhysics ExamplesLearning ResourcesStudy Help