Definition of Magnetic Circuit | What is Magnetic Circuit ? | Magnetic circuit Basics
A Magnetic circuit is defined as the route or path which is followed by magnetic flux. The law of magnetic circuit is in some manner similar to those of the electrical circuit.
Let's consider a solenoid or toroidal iron ring having a magnetic path of l meter, area cross-section A m² and coil of N turns carrying 'I' amperes as shown in below figure.
 |
| Magnetic circuit |
The field strength inside the solenoid is,
H = (NI / l) AT/m
Now, B = μ₀ μᵣ H = [ ( μ₀ μᵣ N I) / l ] Wb/m²
Total flux produce, ϕ = B x A = [ ( μ₀ μᵣ A N I) / l ] Wb
∴ ϕ = [ N I / (l / μ₀ μᵣ A) ] Wb
The 'Nl' which produces magnetization in the magnetic circuit is known as magnetomotive force (m.m.f). Its unit is ampere-turn (AT). It is analogous to e.m.f in an electric circuit.
The reluctance of the circuit is S = ( l / μA ) and is analogous to resistance in an electric circuit.
∴ `flux = \frac {m.m.f} {reluctance}`
Or `phi =\frac{F}{S}`
The above equation can also be called as "Ohm's Law of Magnetic Circuit". Because it resembles a similar expression in an electric circuits i.e.
`current =\frac{e.m.f} {resistance}`
- Some important definitions of Magnetic circuit,
1. Magnetomotive force (m.m.f): This force drive flux through a magnetic circuit. It is measured in ampere-turns. m.m.f between two points is measured by the work done in joules in carrying a unit magnetic pole from one point to another.
2. Ampere-turns (AT): It is the unit of the m.m.f and is given by the product of the number of turns of a magnetic circuit and the current in amperes in those turns.
3. Reluctance: It is the property of the material which opposes the creation of magnetic flux in it. Its unit is AT/Wb.
`[\reluc\tan ce=\frac{l}{\mu A}\]`
where, μ = μ₀ μᵣ
4. Permeance: It is reciprocal of reluctance. It is similar to conductance in an electric circuit. Its unit is Wb/AT or henry.
5. Reluctivity: It is specific reluctance and equals to resistivity which is 'specific resistance.
- How to find Ampere-turns?
As we know the formula of field strength in magnetic circuit is, H = (NI / l) AT/m or NI = H x l
∴ ampere-turns AT = H x l
(a) Find H for each portion of the composite circuit. For air, H = B / μ₀, otherwise H = B / μ₀ μᵣ.
(b) Find the ampere-turns for each path separately by using the relation AT = H x l.
(c) Add up these ampere-turns to get the total ampere-turns for the entire circuit.
- Comparison Between Magnetic and Electric Circuits.
|
Magnetic Circuit |
Electric Circuit |
||||
|
|
||||
|
|
|
||||
|
M.M.F. (ampere-turns) |
E.M.F (volts) |
||||
|
Flux ϕ (webers) |
Current I (amperes) |
||||
|
Flux density B (Wb/m2) |
Current density (A/m2) |
||||
|
Reluctance S = l / μA |
Resistance R = |
||||
|
Permeance (= 1/ reluctance) |
Conductance (= 1/ resistance) |
||||
|
Reluctivity |
Resistivity |
||||
|
Permeability (= 1/ reluctivity) |
Conductivity (= 1/ resistivity) |
||||
|
Total m.m.f. = ϕS1+ ϕS2 + ϕS3+..... |
Total e.m.f. = IR1 + IR2+ IR3 +...... |
- Parallel Magnetic Circuits:
A parallel magnetic circuit consisting of two parallel magnetic path ACB and ADB acted upon by the same m.m.f. Each magnetic path has an average length of 2 (l₁ + l₂). The flux produced by the coil wound on the central core is divided equally at point A between the two outer parallel paths. The reluctance offered by the two parallel paths is = half the reluctance of each path.
Figure b shows the equivalent electrical circuit where resistance offered to the voltage source is = R||R = R / 2
 |
| Parallel Magnetic circuit |
- Series-Parallel Magnetic Circuits:
The below circuit shows the two parallel magnetic circuits ACB and ACD connected across the common magnetic path AB which contains an air-gap of length lg. As usual, the flux ϕ in the common core is divided equally at point A between the two parallel paths which have equal reluctance. The reluctance of the AB consists of
 |
| Series-Parallel Magnetic Circuit |
(i) Air gap reluctance and
(ii) The reluctance of the central core which comparatively negligible. Hence, the reluctance of the central core AB equals only the air-gap reluctance across which are connected two equal parallel reluctances. Hence, the e.m.f. required for this circuit would be the sum of (a) that required for the air-gap and (b) that required for either two paths(not both).
The total resistance offered to the voltage circuit is = R₁ + R||R =R₁ + R/2.
Post a Comment