22.1 MAGNETIC FLUX
22.1.1 Magnetic Flux
1 Definition
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| Figure 22.1: Shows how magnetic flux depends on the angle between the magnetic field and the normal to the surface. |
Consider a very small flat surface of area A inside a magnetic field, which may not be uniform. If the area is very small, a non-uniform field in the very small region enclosed by the area may be considered to be uniform. Hence, the strength of the field at any point on that surface can be considered to be the same. Let the strength (or magnetic flux density) be of magnitude B at any point on the surface, and let its direction of the field make an angle θ to the normal of the area, as shown in Figure 22.1. Then the amount of magnetic flux Φ existing at the area of the small surface is given by
Φ = BA⊥
where A⊥ is the area of the surface perpendicular to the direction of the field. We have
A⊥ = A cos θ
Hence, magnetic flux is also expressed as
Φ = BA cos θ
Notice that
(a) if θ = 0°, the surface is perpendicular to the field, and Φ = BA, which is a maximum value.
(b) if θ = 90°, the surface is parallel to the field, and Φ = 0
2 Unit
The SI unit for magnetic flux is the weber (Wb) or T m2.
3 Quantity
Magnetic flux is a scalar quantity.
EXAMPLE 22.1
A small surface of area 10 mm2 inside a uniform magnetic field of strength 0.50 T is inclined at an angle of α to the direction of the field. Determine the magnetic flux through the surface if
(a) α = 0° (b) α = 30° (c) α = 90°
Answer
(a) α = 0°:
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| Figure 22.2 Illustrates the relationship between α and θ, where α + θ = 90°. |
Refer to Figure 22.2. Notice that the angle α between the area and the direction of the magnetic field is related to angle θ by the expression
α = 90° − θ
Hence, we have
Φ = BA cos θ
= BA cos (90° − α)
= BA sin α
= BA sin 0 = 0
If the area is parallel to the direction of the magnetic field (α = 0), the magnetic flux is zero.
(b) α = 30°:
Φ = BA sin α
= (0.50)(10 × 10−6) sin 30° = 2.5 × 10−6 Wb
(c) α = 90°:
Φ = (0.50)(10 × 10−6) sin 90° = 5.0 × 10−6 Wb
If the area is perpendicular to the direction of the magnetic field (α = 90°), the magnetic flux is a maximum.
22.1.2 Model for Magnetic Flux
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| Figure 22.3 Shows how magnetic flux varies with field line density across equal surface areas. |
- Figure 22.3 shows a portion of a magnetic field represented by a set of magnetic field lines. There are two very small areas, X and Y, of the same size placed roughly normal to the field lines.
- The number of magnetic field lines passing through an area reflects the magnitude of the magnetic flux through that area. The larger the number of field lines passing through an area, the greater would be the magnetic flux through that area. For example, the magnetic flux through X is greater than that through Y because more field lines pass through X than through Y.
22.1.3 Magnetic Flux Linkage
Refer to Figure 22.3. Suppose that area X is replaced by a coil of N turns. Now we have magnetic field lines passing through the N turns of the coil, giving the impression that the field lines are ‘linking up’ all the turns of the coil. This magnetic flux linkage is given by
flux linkage = NΦ
or,
flux linkage = NBA cos θ
where A is the area of the plane of the coil. If the coil is circular and has diameter d, then we have
A = 1/4 πd2
EXAMPLE 22.2
A circular coil has 10 turns and diameter 2.0 cm. It is placed in a uniform magnetic field of strength 300 mT. The plane of the coil and the direction of the field make an angle of 30°. Determine the magnetic flux linkage through the coil.
Answer
Given N = 10, d = 2.0 × 10−2 m, B = 0.300 T, α = 30°.
flux linkage = NBA cos θ
= NBA sin α
= NB (1/4 πd2) sin α
= (10)(0.300)(1/4 π)(2.0 × 10−2)2 sin 30°
= 4.7 × 10−4 Wb
EXAMPLE 22.3
A circular coil is placed in a uniform magnetic field with its plane perpendicular to the direction of the field. Suggest ways which can be employed to reduce the magnetic flux linkage through the coil by 20%.
Answer
(a) Changing magnetic flux density
The initial magnetic flux linkage Φ0 is given by
Φ0 = NBA cos 0°
Keep N and A constant. The final flux linkage Φ is given by
Φ = NBA
Reduction of flux linkage
ΔΦ / Φ0 = (Φ − Φ0) / Φ0
= (NBA − NB0A) / NB0A
= B / B0 − 1 = −20%
B / B0 = 1 − 0.2 = 0.8
B = 0.8B0
Use a magnetic field of magnetic flux density 0.8B0.
(b) Changing size of coil
Keep N and B constant. We have
ΔΦ / Φ0 = (NBA − NBA0) / NBA0
= A / A0 − 1
= d2 / d02 − 1 = −20%
(A = 1/4 πd2)
where d and d0 are the final and initial diameters of the coil.
d / d0 = √(1 − 0.20) = 0.89
d = 0.89d0
Use a coil whose diameter is 0.89d0.
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| Figure 22.4 Shows how rotating a coil in a magnetic field reduces the flux linkage. |
(c) Changing angle between plane of coil and direction of magnetic field
Keep N, B and A constant. Rotate the coil through an angle of θ about an axis which lies on the plane of the coil, passes through the centre of the coil and is perpendicular to the direction of the field until the flux linkage is reduced by 20%, as shown in Figure 22.4.
ΔΦ / Φ0 = (NBA cos θ − NBA cos 0°) / NBA cos 0°
= cos θ − 1 = −0.2
cos θ = 1 − 0.2 = 0.8
θ = 36.9°
θ = 36.9° to the direction of the field
EXAMPLE 22.4
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| Figure 22.5 Shows a circular coil wound around a solenoid where magnetic flux linkage is produced by the current in the solenoid. |
A circular coil of 20 turns and diameter 10 cm is wound tightly round the middle portion of a long solenoid, as shown in Figure 22.5. A current of 2.0 A flows through the solenoid, which has 500 turns per metre. Determine the magnetic flux linkage through the coil.
Answer
Solenoid: The magnetic flux density B inside the middle portion of the solenoid produced by current I is given by
B = μ0nI
= (4π × 10−7)(500)(2) = 1.26 × 10−3 T
Coil: Magnetic flux linkage Φ through the coil is given by
Φ = NB(1/4 πd2) cos 0°
= (20)(1.26 × 10−3)(1/4 π)(0.102)
= 2.0 × 10−4 Wb
EXERCISE 22.1
A rectangular coil has 100 turns and dimensions of 2.0 cm × 3.0 cm. It is placed in a uniform magnetic field of strength 200 mT. The plane of the coil and the direction of the field make an angle of 30°. Determine the magnetic flux linkage through the coil.
22.1.4 Change of Flux Linkage
Flux linkage is given by
flux linkage = NBA cos θ
Hence, its magnitude will change if N, B, A and θ change. In practice, we normally do not change the number of turns N and the size (area A) of the coil. If we wish to change the magnetic flux linkage, we may have to change B and/or θ.
B and θ may be changed in the following ways:
1 Changing B:
The magnetic field strength may be changed in two ways:
(a) Move the coil to another place where the magnetic strength is different. For example, a coil initially in position X in Figure 22.3 has 7 field lines linking the coil. If it is moved to position Y, only 5 field lines pass through it.
(b) The magnetic field strength B at a point may be changing with time t. For example, we may have
B = k sin ωt
2 Changing θ:
Angle θ can be changed by rotating the coil about an axis which passes through the centre of the coil and lies on the plane of the coil. A coil rotating in a dynamo or a motor is one such example.
EXAMPLE 22.5
The area of the plane of a circular coil with 10 turns is 2.0 × 10−3 m2. It is placed in a magnetic field of strength Bi = 200 mT whose direction makes an angle αi = 20° with the plane of the coil. Determine the change of flux linkage if
(a) the magnetic field strength changes to Bf = 500 mT
(b) the angle α changes to αf = 50° by rotating the coil.
Answer
(a)
initial flux linkage = N Bi A sin αi
final flux linkage = N Bf A sin αi
change of flux linkage = N Bf A sin αi − N Bi A sin αi
= N A sin αi (Bf − Bi)
= (10)(2.0 × 10−3)(sin 20°)(500 − 200) × 10−3
= 2.1 × 10−3 Wb
(b)
initial flux linkage = N Bi A sin αi
final flux linkage = N Bi A sin αf
change of flux linkage = N Bi A sin αf − N Bi A sin αi
= N Bi A (sin αf − sin αi)
= (10)(200 × 10−3)(2.0 × 10−3)(sin 50° − sin 20°)
= 1.7 × 10−3 Wb
Applications of Magnetic Flux and Flux Linkage
Magnetic flux and flux linkage are widely used in many electrical and electronic devices. The table below summarizes key applications and their functions.
| No | Application | Description |
|---|---|---|
| 1 | Electric Generators | Convert mechanical energy into electrical energy through changing magnetic flux linkage, producing emf based on Faraday’s Law. |
| 2 | Electric Motors | Use magnetic fields and electric current to generate force and produce rotational motion. |
| 3 | Transformers | Transfer electrical energy between coils by changing magnetic flux in the core, inducing voltage in the secondary coil. |
| 4 | Inductors | Store energy in the form of magnetic flux and resist changes in current through self-induction. |
| 5 | Magnetic Sensors | Detect magnetic fields and are used in smartphones, automotive systems, and industrial devices. |
| 6 | Wireless Charging | Transfer energy wirelessly using changing magnetic flux between coils via electromagnetic induction. |
| 7 | Dynamos | Generate electricity by continuously changing magnetic flux linkage in a rotating coil. |
Conclusion
| Key Point | Summary |
|---|---|
| Definition | Magnetic flux and flux linkage describe how magnetic fields interact with surfaces and coils. |
| Importance | They are essential for understanding the generation, transfer, and use of electrical energy. |
| Applications | Used in generators, transformers, motors, wireless charging, and magnetic sensors. |
| Key Formula | Magnetic flux (Φ = BA cos θ) and flux linkage (NΦ) are fundamental in calculations. |
| Real-World Impact | These concepts enable the design of efficient electrical and electronic systems. |
| Learning Benefit | Understanding these concepts improves problem-solving skills in physics and engineering. |
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