(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant 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

3. 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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# Homework Help: C core help!

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