12.10 Magnetic Flux Linkage | Electromagnetism | Physics 11

Magnetic flux linkage is the total magnetic flux that passes through all the turns of a coil or solenoid. It is calculated as the product of the magnetic flux through a single turn and the total number of turns in the coil. This concept is crucial for understanding electromagnetic induction in multi-turn devices.

Magnetic Flux Linkage Formula

The magnetic flux linkage ( $Φ_{N}$ ) is defined as:

$Φ_{N} = N Φ$

Since the magnetic flux ( $Φ$ ) through a single turn is given by $Φ = B A cos θ$ , the general formula for flux linkage is:

$Φ_{N} = N B A cos θ$

Where:

$Φ_{N}$ is the magnetic flux linkage
$N$ is the number of turns in the coil (dimensionless)
$B$ is the magnetic flux density (T)
$A$ is the cross-sectional area of the coil (m²)
$θ$ is the angle between the magnetic field $B$ and the normal to the coil's plane

When the magnetic field is perpendicular to the plane of the coil (i.e., parallel to the area normal, $θ = 0^{\circ}$ ), $cos (0^{\circ}) = 1$ and flux linkage is maximum:

$Φ_{N} = N B A$

When the field is parallel to the plane of the coil ( $θ = 9 0^{\circ}$ ), $cos (9 0^{\circ}) = 0$ and flux linkage is zero.

Units of Magnetic Flux Linkage

The unit for magnetic flux linkage is the Weber turn (Wb turn). Since the number of turns $N$ is dimensionless, the Weber turn is dimensionally equivalent to the Weber (Wb), or equivalently $T \cdot m^{2}$ .

Quantity	Symbol	Unit
Magnetic Flux Linkage	$Φ_{N}$	Weber turns (Wb turns)
Number of Turns	$N$	dimensionless
Magnetic Flux	$Φ$	Weber (Wb)
Magnetic Flux Density	$B$	Tesla (T)
Area	$A$	m²

The dimensions of magnetic flux linkage are: $[Φ_{N}] = [M L^{2} T^{- 2} A^{- 1}]$

Difference Between Magnetic Flux and Magnetic Flux Linkage

	Magnetic Flux ( $Φ$ )	Magnetic Flux Linkage ( $Φ_{N}$ )
Definition	Field through a single loop	Field through all $N$ turns
Formula	$Φ = B A cos θ$	$Φ_{N} = N Φ = N B A cos θ$
Unit	Weber (Wb)	Weber turn (Wb turn)

Example Calculations

Example 1: Solenoid in a Magnetic Field

A solenoid with 300 turns and a cross-sectional area of $0.80 m^{2}$ is placed perpendicular to a magnetic field with a flux density of 4 mT.

Given:

$N = 300$
$A = 0.80 m^{2}$
$B = 4 mT = 4 \times 1 0^{- 3} T$
$θ = 0^{\circ}$ (field perpendicular to plane, i.e., along normal)

Solution:

$Φ_{N} = N B A = (300) (4 \times 1 0^{- 3}) (0.80) = 0.96 Wb turns$

Example 2: Another Solenoid Calculation

A solenoid has 200 turns, a cross-sectional area of $0.004 m^{2}$ , and is in a magnetic field with a flux density of 12.0 mT.

Given:

$N = 200$
$A = 0.004 m^{2}$
$B = 12.0 mT = 12 \times 1 0^{- 3} T$

Solution:

$Φ_{N} = N B A = (200) (12 \times 1 0^{- 3}) (0.004) = 9.6 \times 1 0^{- 3} Wb turns$

Key Formulae Summary

Condition	Formula
General (angle $θ$ to normal)	$Φ_{N} = N B A cos θ$
Field along normal ( $θ = 0^{\circ}$ , maximum)	$Φ_{N} = N B A$
Field in plane of coil ( $θ = 9 0^{\circ}$ , zero)	$Φ_{N} = 0$

Understanding flux linkage is essential for determining the induced voltage in electromagnetic systems — it is the foundational quantity in transformers, inductors, and electric generators.

Magnetic Flux Linkage Formula

The magnetic flux linkage ( $Φ_{N}$ ) is defined as:

$Φ_{N} = N Φ$

Since the magnetic flux ( $Φ$ ) through a single turn is given by $Φ = B A cos θ$ , the general formula for flux linkage is:

$Φ_{N} = N B A cos θ$

Where:

$Φ_{N}$ is the magnetic flux linkage
$N$ is the number of turns in the coil (dimensionless)
$B$ is the magnetic flux density (T)
$A$ is the cross-sectional area of the coil (m²)
$θ$ is the angle between the magnetic field $B$ and the normal to the coil's plane

When the magnetic field is perpendicular to the plane of the coil (i.e., parallel to the area normal, $θ = 0^{\circ}$ ), $cos (0^{\circ}) = 1$ and flux linkage is maximum:

$Φ_{N} = N B A$

When the field is parallel to the plane of the coil ( $θ = 9 0^{\circ}$ ), $cos (9 0^{\circ}) = 0$ and flux linkage is zero.

Units of Magnetic Flux Linkage

Quantity	Symbol	Unit
Magnetic Flux Linkage	$Φ_{N}$	Weber turns (Wb turns)
Number of Turns	$N$	dimensionless
Magnetic Flux	$Φ$	Weber (Wb)
Magnetic Flux Density	$B$	Tesla (T)
Area	$A$	m²

The dimensions of magnetic flux linkage are: $[Φ_{N}] = [M L^{2} T^{- 2} A^{- 1}]$

Difference Between Magnetic Flux and Magnetic Flux Linkage

	Magnetic Flux ( $Φ$ )	Magnetic Flux Linkage ( $Φ_{N}$ )
Definition	Field through a single loop	Field through all $N$ turns
Formula	$Φ = B A cos θ$	$Φ_{N} = N Φ = N B A cos θ$
Unit	Weber (Wb)	Weber turn (Wb turn)

Example Calculations

Example 1: Solenoid in a Magnetic Field

A solenoid with 300 turns and a cross-sectional area of $0.80 m^{2}$ is placed perpendicular to a magnetic field with a flux density of 4 mT.

Given:

$N = 300$
$A = 0.80 m^{2}$
$B = 4 mT = 4 \times 1 0^{- 3} T$
$θ = 0^{\circ}$ (field perpendicular to plane, i.e., along normal)

Solution:

$Φ_{N} = N B A = (300) (4 \times 1 0^{- 3}) (0.80) = 0.96 Wb turns$

Example 2: Another Solenoid Calculation

A solenoid has 200 turns, a cross-sectional area of $0.004 m^{2}$ , and is in a magnetic field with a flux density of 12.0 mT.

Given:

$N = 200$
$A = 0.004 m^{2}$
$B = 12.0 mT = 12 \times 1 0^{- 3} T$

Solution:

$Φ_{N} = N B A = (200) (12 \times 1 0^{- 3}) (0.004) = 9.6 \times 1 0^{- 3} Wb turns$

Key Formulae Summary

Condition	Formula
General (angle $θ$ to normal)	$Φ_{N} = N B A cos θ$
Field along normal ( $θ = 0^{\circ}$ , maximum)	$Φ_{N} = N B A$
Field in plane of coil ( $θ = 9 0^{\circ}$ , zero)	$Φ_{N} = 0$

Understanding flux linkage is essential for determining the induced voltage in electromagnetic systems — it is the foundational quantity in transformers, inductors, and electric generators.