How to Calculate Flux Density in a C-Core with an Air Gap?

In summary, flux density is a measure of the strength of a magnetic field, calculated by dividing the magnetic flux by the area it passes through. It is measured using a gaussmeter and is affected by the strength of the magnetic field, the size of the area, and the alignment. Flux density is important in understanding magnetic fields and is used in various technologies. It is also used in research and experiments to study the properties of materials and their interactions with magnetic fields.
  • #1
Sanna Khan
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Homework Statement



A "c" core has a 500 turn coil carrying 1.5 A, wound around it and a conductor carrying 25A is placed in the air gap.
If the air gap has the dimensions:
L= 125mm
D= 20mm
and a reluctance of 250,000 at/wb determine
a) the flux
B) the flux density
c) the force exerted on the conductor.

Please can someone check to see if my working out is correct or not?

many thanx

Homework Equations



Applying Ampere's circuit law around the circumference of
the ring:

∫ H.dL = NIc
Since H is constant along the circumference and is parallel to dl
H ∫dL = NIc

H = NI/L

B = μNI/L

In general, the magnetic flux for the ring is:

Φ = BA = μNIA/L = NI / (L/μA) = NI/R,

where,
reluctance: R = L/μA

But since we have a 'c' ring there is an air gap.

The flux will then be given by:

Φ = NI / (R + Ro) = NI / (R + Lo/μoA)

where,
reluctance: R = 250,000 At/wb
magnetic field intensity: H
magnetic field density: B
permeability: μ
free space permeability: μo = 4πx10^-7 H/m
area: A = π 0.125²/4 = π = π3.910^-³ m²
turns: N = 500
current: I = 1.5 A
air gap length: Lo = 0.125 m
air gap width: Do = 0.02
reluctance: R = 0.125/(μπ10^-4) = 250000 At/wb

magnetic flux: Φ

Φ = NI / (R + Lo/μoA)

Φ = 500*1.5/{250000 + (0.02)/[(4πx10^-7)π3.9x10^-³]}

Φ = 750/[(2.5 + 12.9)x10^5] = 4.87.x10^-4 wb


b) flux density: B

B = Φ/A = 4.87x10-4/ π3.9x10^-³ = 0.4 Wb/m²


c) magnetomotive force: F

F = NI = 500*25 = 12500 N



The Attempt at a Solution



magnetic flux: Φ

Φ = NI / (R + Lo/μoA)

Φ = 500*1.5/{250000 + (0.02)/[(4πx10^-7)π3.9x10^-³]}

Φ = 750/[(2.5 + 12.9)x10^5] = 4.87.x10^-4 wb

b) flux density: B

B = Φ/A = 4.87x10-4/ π3.9x10^-³ = 0.4 Wb/m²

c) magnetomotive force: F

F = NI = 500*25 = 12500 N
 
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  • #2


Your working out looks correct. However, I would suggest using proper units in your calculations to avoid confusion. For example, instead of using "At/wb" for reluctance, use "A/Wb" or "A/Tm". Also, make sure to include the units in your final answers. For example, the flux density should be 0.4 Wb/m², not just 0.4. Other than that, your calculations and equations seem correct. Good job!
 

What is flux density?

Flux density, also known as magnetic flux density or magnetic induction, is a measure of the strength of a magnetic field. It represents the amount of magnetic flux passing through a unit area.

How is flux density measured?

Flux density is measured using a device called a gaussmeter, which measures the strength of a magnetic field in gauss or tesla units. It can also be calculated by dividing the magnetic flux by the area it passes through.

What factors affect flux density?

The strength of flux density is affected by the strength of the magnetic field, the size of the area it passes through, and the angle at which the field and area are aligned.

Why is flux density important?

Flux density is an important concept in understanding the behavior of magnetic fields and their effects on materials. It is also used in various technologies, such as motors, generators, and transformers.

How is flux density used in research and experiments?

Flux density is used in many scientific experiments and research studies to measure and manipulate magnetic fields. It is also used to study the properties of materials and their interactions with magnetic fields.

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