Question

An iron ring has a mean diameter of 15 cm, a cross-section of 20 cm2 and a radial gap of 0·5 mm cut in it. It is uniformly wound with 1500 turns of insulated wire and a magnetizing current of 1 A produces a flux of 1 mWb. Neglecting the effect of magnetic leakage and fringing, calculate (i) reluctance of the magnetic circuit, (ii) relative permeability of iron​

Answer 1

Answer:

Given

L = 0.4712 m = 3.14d

A = 20 10-4 m2

Lg = 0.5 mm

N = 1500

I = 1 A

Solution

a) the reluctance of the magnetic circuit

R = 1500/(110-3)

= 1.510-6 A/wb expression 2

b) the relative permeability

 

ur = 143.9

Explanation:

Answer 2

Answer:

(i) Reluctance of the magnetic circuit is 1.510-6 A/wb

(ii) Relative permeability of iron​ is μr = 143.9

Explanation:

Given

L = 0.4712

m = 3.14 d

A = 20× (10-4) m²

Lg = 0.5 mm

N = 1500

I = 1 A

Neglecting the effect of magnetic leakage and fringing

a) The reluctance of the magnetic circuit is Magnetomotive force (mmf) to magnetic flux ratio is used to measure the magnetic circuit's resistance. Depending on the geometry and composition of an object, it stands for the opposition to magnetic flow.

R = 1500÷(110-3)

R= 1.510-6 A/wb

b) relative permeability is the ratio of the permeability to a certain fluid in the presence of additional fluids to the absolute permeability known as relative permeability.

μr = 143.9

Therefore the reluctance of the magnetic circuit is 1.510-6 A/wb and The relative permeability is μr = 143.9

To learn more about magnetic circuits, visit:

https://brainly.in/question/641926

To learn more about magnetic leakage and fringing, visit:

https://brainly.in/question/37795156

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