An Iron Core's Magnetisation

solved Magnetisation

Q. A solenoid has a core of a material with relative permeability 400 $μ_{r} = 400$ . The windings of the solenoid are insulated from the core and carry a current of 2 A $I = 2$ A. If the number of turns is 1000 m⁻¹ $n = 1000$ m⁻¹, what is the magnetisation M of the core, in A/m?

Inputs

permeability

μ_{r}

400

current

I

2 A

turns

n

1000 m⁻¹

Outputs

magnetisation

M

How it's solved

Solved with Magnetisation

Given

$μ_{r} = 400$
$I = 2$ A
$n = 1000$ m⁻¹

Find $M$ .

Working

intensity of a solenoid(core) $H = n \cdot I$
$H = 1000 \cdot 2$
$H = 2000$ A/m
magnetisation of a core(core) $M = (μ_{r} - 1) \cdot H$
$M = (400 - 1) \cdot 2000$
$M = 798000$ A/m
$M \approx 8.0 \times 10^{5}$ A/mto 2 significant figures

Conclusion

The core's own atomic magnets line up with H and add far more magnetisation than the windings make alone.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :an_iron_core_s_magnetisation do4  kind :numerical5  describes "A solenoid has a core of a material with relative permeability %{permeability}. The windings of the " \6            "solenoid are insulated from the core and carry a current of %{current}. If the number of turns is " \7            "%{turns}, %{magnetisation}, in A/m?"8  given permeability: "400"9  given current: "2 A"10  given turns: "1000 m-1"11  asks magnetisation: "what is the magnetisation M of the core"12end

solution.rb

1require_relative "../questions/an_iron_core_s_magnetisation"2require_relative "../science/magnetism_and_matter/magnetisation"34solve :an_iron_core_s_magnetisation, using: Magnetisation do5  objects :core67  take permeability: core.relative_permeability8  take current: core.current9  take turns: core.turns_per_length1011  magnetisation = core.magnetisation12  answer magnetisation:, in: "A/m", figures: 21314  conclusion "The core's own atomic magnets line up with H and add far more magnetisation than the windings make alone."15end

playground.rb

1require_relative "../solutions/an_iron_core_s_magnetisation"23play_with :an_iron_core_s_magnetisation do4  title "An Iron Core's Magnetisation"5end

The laws it drew on

quantity	written	units	within	from
permeability of free space	$μ_{0}$	N/A²	—	Magnetisation
relative permeability	$μ_{r}$	—	—	Magnetisation
current	$I$	A	—	Magnetisation
turns per length	$n$	m⁻¹	—	Magnetisation
magnetic intensity	$H$	A/m	—	Magnetisation
magnetisation	$M$	A/m	—	Magnetisation
magnetic field	$B$	T	—	Magnetisation
magnetising current	$I_{m}$	A	—	Magnetisation
magnetic intensity of ground	$H_{g}$	A/m	—
turns per length of ground	$n_{g}$	m⁻¹	—
current of ground	$I_{g}$	A	—
magnetic intensity of core	$H$	A/m	—
turns per length of core	$n$	m⁻¹	—
current of core	$I$	A	—
magnetisation of ground	$M_{g}$	A/m	—
relative permeability of ground	$μ_{r,g}$	—	—
magnetisation of core	$M$	A/m	—
relative permeability of core	$μ_{r}$	—	—
magnetic field of ground	$B_{g}$	T	—
magnetic field of core	$B$	T	—
magnetising current of ground	$I_{m,g}$	A	—
magnetising current of core	$I_{m}$	A	—

H_{c} = n_{c} \cdot I_{c}

intensity of a solenoid

M_{c} = (μ_{r,c} - 1) \cdot H_{c}

magnetisation of a core

B_{c} = μ_{r,c} \cdot μ_{0} \cdot H_{c}

field in a magnetised core

B_{c} = μ_{0} \cdot n_{c} \cdot (I_{c} + I_{m,c})

the current that would do without the core

H_{g} = n_{g} \cdot I_{g}

intensity of a solenoid(ground)

H = n \cdot I

intensity of a solenoid(core)

M_{g} = (μ_{r,g} - 1) \cdot H_{g}

magnetisation of a core(ground)

M = (μ_{r} - 1) \cdot H

magnetisation of a core(core)

B_{g} = μ_{r,g} \cdot μ_{0} \cdot H_{g}

field in a magnetised core(ground)

B = μ_{r} \cdot μ_{0} \cdot H

field in a magnetised core(core)

B_{g} = μ_{0} \cdot n_{g} \cdot (I_{g} + I_{m,g})

the current that would do without the core(ground)

B = μ_{0} \cdot n \cdot (I + I_{m})

the current that would do without the core(core)