Magnetic Flux Calculation: Coil With Two Sections

by Admin 50 views
Magnetic Flux Calculation: Coil with Two Sections

Hey physics enthusiasts! Today, we're diving deep into the fascinating world of electromagnetism to tackle a problem involving a coil with N = 200 turns. We've got a coil that's not uniform; it's split into two sections, or tramos, each with its own characteristics. We'll be calculating the magnetic flux, a fundamental concept that tells us about the magnetic field passing through a given area. This is super important in understanding how inductors, transformers, and many other electrical components work, guys. So, let's break down this problem step by step and make sure we nail this calculation!

Understanding the Setup: A Tale of Two Sections

First off, let's get a clear picture of our setup. We have a coil with a total of N = 200 espiras (turns). This coil isn't a simple cylinder; it's divided into two distinct sections, Tramo 1 and Tramo 2. Each section has its own cross-sectional area (S) and length (L).

  • Tramo 1: This section has a cross-sectional area of S1 = 30 cm² and a length of L1 = 20 cm.
  • Tramo 2: This section is a bit larger, with a cross-sectional area of S2 = 35 cm² and a length of L2 = 40 cm.

We're also given some crucial information about the material the coil is wound around. The relative permeability (μr) of the material is a whopping 2500. This value tells us how much better the material is at supporting the formation of a magnetic field compared to a vacuum. A high relative permeability means the material can concentrate magnetic field lines effectively.

Finally, we know the magnetic induction (B) in Tramo 2 is 0.6 T (Tesla). Magnetic induction is essentially the strength of the magnetic field. We need to use all this info to calculate the magnetic flux.

Diving into Magnetic Flux: The Core Concept

So, what exactly is magnetic flux, anyway? In simple terms, magnetic flux (Φ) is a measure of the total magnetic field that passes through a given surface. Think of it like counting the number of magnetic field lines going through a loop of wire. The more field lines that pass through, the greater the magnetic flux. Mathematically, for a uniform magnetic field passing perpendicularly through a surface, it's given by:

Φ = B * A

Where:

  • Φ is the magnetic flux (measured in Webers, Wb)
  • B is the magnetic field strength (or magnetic induction, measured in Tesla, T)
  • A is the area through which the magnetic field passes (measured in square meters, m²)

However, our problem is a bit more complex because the coil has two sections with different cross-sectional areas and potentially different magnetic field strengths. This means we can't just use a single formula for the entire coil. We need to consider each section separately. Also, remember that the magnetic field lines might not be perpendicular to the surface. In a more general case, it's a dot product: Φ = ∫ B ⋅ dA.

Calculating Magnetic Flux for Each Section

Let's get down to business and calculate the magnetic flux for each section of our coil. Remember, we need to be consistent with our units, so we'll convert everything to meters and square meters.

Tramo 1:

  • Area (S1) = 30 cm² = 30 * (10⁻² m)² = 30 * 10⁻⁴ m² = 0.003 m²
  • Length (L1) = 20 cm = 0.20 m

We know the magnetic induction in Tramo 2 (B2 = 0.6 T). To find the flux in Tramo 1, we first need to determine the magnetic field strength (B1) in Tramo 1. The magnetic field inside a solenoid or a coil is related to the number of turns per unit length and the current. However, we aren't given the current directly. Instead, we can think about how the magnetic field is distributed. In a uniform core material, the magnetic flux should be continuous. This means the flux passing through Tramo 1 must be equal to the flux passing through Tramo 2, assuming the magnetic field lines are contained within the core. Therefore, the magnetic flux in Tramo 1 is:

Φ₁ = B₁ * S₁

And the magnetic flux in Tramo 2 is:

Φ₂ = B₂ * S₂

Since the magnetic flux density (B) is related to the magnetomotive force (MMF) and reluctance (ℛ) by B = MMF / (ℛ * A), and MMF is constant throughout the coil (assuming a single current path), and the flux (Φ = B*A) is also constant if the core is continuous and no flux is lost, we can equate the flux through both sections. Therefore, Φ₁ = Φ₂.

Let's calculate the flux in Tramo 2 first, as we have all the necessary information:

  • B2 = 0.6 T
  • S2 = 35 cm² = 35 * 10⁻⁴ m² = 0.0035 m²

Φ₂ = B₂ * S₂ = 0.6 T * 0.0035 m² = 0.0021 Wb

Since the magnetic flux is conserved within the core material (assuming no leakage), the magnetic flux in Tramo 1 must be the same as in Tramo 2.

Φ₁ = Φ₂ = 0.0021 Wb

So, the magnetic flux through both sections of the coil is 0.0021 Webers. Pretty neat, right? This means that even though the cross-sectional areas are different, the total magnetic field lines passing through are the same, which is a key concept in magnetic circuits.

Calculating the Magnetic Field in Tramo 1

Now that we know the magnetic flux is the same for both sections, we can calculate the magnetic field strength (B1) in Tramo 1. We know:

  • Φ₁ = 0.0021 Wb
  • S₁ = 30 cm² = 0.003 m²

Using the formula Φ₁ = B₁ * S₁, we can rearrange it to solve for B1:

B₁ = Φ₁ / S₁

B₁ = 0.0021 Wb / 0.003 m² = 0.7 T

So, the magnetic induction in Tramo 1 is 0.7 Tesla. Notice how the magnetic field strength is higher in Tramo 1 where the cross-sectional area is smaller. This makes sense because the same amount of magnetic flux has to pass through a narrower area, concentrating the field lines more intensely. It's like squeezing water through a narrower pipe – the pressure (analogous to magnetic field strength) increases.

Calculating the Reluctance of Each Section

Reluctance (ℛ) is the magnetic equivalent of electrical resistance. It's a measure of how much a material opposes the establishment of a magnetic field. The formula for reluctance is:

ℛ = L / (μ * A)

Where:

  • L is the length of the magnetic path (in meters)
  • μ is the permeability of the material (in H/m)
  • A is the cross-sectional area (in m²)

The permeability (μ) of a material is given by μ = μr * μ₀, where μr is the relative permeability and μ₀ is the permeability of free space (approximately 4π * 10⁻⁷ H/m).

Let's calculate the reluctance for each section:

Tramo 1:

  • L₁ = 0.20 m
  • S₁ = 0.003 m²
  • μ = μr * μ₀ = 2500 * (4π * 10⁻⁷ H/m) ≈ 0.00314 H/m

ℛ₁ = L₁ / (μ * S₁) = 0.20 m / (0.00314 H/m * 0.003 m²) ≈ 0.20 / 0.00000942 ≈ 21231.42 H⁻¹

Tramo 2:

  • L₂ = 0.40 m
  • S₂ = 0.0035 m²
  • μ = 0.00314 H/m (same material)

ℛ₂ = L₂ / (μ * S₂) = 0.40 m / (0.00314 H/m * 0.0035 m²) ≈ 0.40 / 0.00001099 ≈ 36396.72 H⁻¹

As you can see, Tramo 2 has a higher reluctance than Tramo 1. This is due to its longer length and larger cross-sectional area, though the length plays a more significant role here in increasing the opposition to the magnetic field.

Calculating the Total Reluctance

Since both sections are in series within the magnetic circuit, their reluctances add up. The total reluctance (ℛ_total) of the coil is:

ℛ_total = ℛ₁ + ℛ₂

ℛ_total ≈ 21231.42 H⁻¹ + 36396.72 H⁻¹ ≈ 57628.14 H⁻¹

This total reluctance tells us the overall resistance of the entire coil to the magnetic field.

Calculating the Magnetomotive Force (MMF)

The magnetomotive force (MMF) is the driving force behind the magnetic field in a magnetic circuit, analogous to electromotive force (EMF) in an electrical circuit. It's given by:

MMF = N * I

Where:

  • N is the total number of turns (200)
  • I is the current flowing through the coil (which we don't know yet).

However, we can also relate MMF to the total flux and total reluctance:

MMF = Φ_total * ℛ_total

We know the total flux (Φ_total = 0.0021 Wb, as it's the same for both sections) and the total reluctance (ℛ_total ≈ 57628.14 H⁻¹).

MMF ≈ 0.0021 Wb * 57628.14 H⁻¹ ≈ 121.02 Amperes-turns

This MMF is the effective driving force created by the coil to establish the magnetic field.

Calculating the Current (I)

Now we can finally calculate the current (I) flowing through the coil using the MMF and the total number of turns (N).

MMF = N * I

Rearranging for I:

I = MMF / N

I ≈ 121.02 Amperes-turns / 200 turns ≈ 0.605 Amperes

So, the current flowing through the coil is approximately 0.605 Amperes. This is the current needed to produce the given magnetic field strength in Tramo 2!

Summary of Results

Let's recap our findings:

  • Magnetic Flux (Φ₁ and Φ₂): 0.0021 Wb
  • Magnetic Induction in Tramo 1 (B₁): 0.7 T
  • Reluctance of Tramo 1 (ℛ₁): ≈ 21231.42 H⁻¹
  • Reluctance of Tramo 2 (ℛ₂): ≈ 36396.72 H⁻¹
  • Total Reluctance (ℛ_total): ≈ 57628.14 H⁻¹
  • Magnetomotive Force (MMF): ≈ 121.02 Amperes-turns
  • Current (I): ≈ 0.605 Amperes

This was a pretty involved calculation, guys, but we managed to break it down piece by piece. Understanding magnetic flux, reluctance, and MMF is key to mastering electromagnetism. Keep practicing these concepts, and you'll be acing your physics problems in no time! Stay curious and keep exploring the amazing world of physics!