Magnetic Energy of Solenoid With/Without Core

In summary, the formula for magnetic field energy density is related to the permeability of the material being used in the solenoid. An air-filled core solenoid will store more energy than a steel core solenoid due to the smaller denominator in the formula. However, for the same magnitude of B, the energy density will be higher in the case of the steel core solenoid. This is because it takes less effort to increase the magnitude of B in a steel core solenoid, resulting in less potential energy being stored. There may be instances where the same magnitude of B is achieved in both solenoids, but the energy density will still be higher in the steel core solenoid.
  • #1
dgreenheck
23
0
I have a question about solenoids.

The formula for the magnetic field energy density is:

[itex]\frac{1}{2}\frac{B^{2}}{μ}[/itex]

If I have an air-filled core, then μ=μ0. If I have a steel core, then μ will be ~ 100μ0. This implies that an air-filled core solenoid stores more energy than a steel core solenoid since the denominator for the air-core case is smaller.

Doesn't this contradict the fact that solenoids with a ferromagnetic core produce stronger magnetic fields than a solenoids with no core?
 
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  • #2
In a solenoid B=μnI where n is the number of turns per unit length, so your value for 'B' will increase with a change in the medium.
 
  • #3
dgreenheck said:
This implies that an air-filled core solenoid stores more energy than a steel core solenoid since the denominator for the air-core case is smaller.
It stores more energy for the same magnitude of B. You have:

Uair = 1/2*B2*V/μair
Usteel = 1/2*B2*V/μsteel = 1/2*B2*V/(100*μair) ⇔
Uair = 100*Usteel

where U is the potential energy stored by the solenoid and V is the volume of its interior.

Think of it this way, it takes less effort (work) to increase the magnitude of B in a steel core solenoid than in an air core solenoid, so for the same magnitude of B, less potential energy is stored.
 
  • #4
Okay, both of your responses make sense. Is there ever a time where you would have the same B, though? If the B field is always dependent on the permeability of the material it's flowing through, the squared factor of mu on top will cancel out the factor of mu in the denominator and the energy density will always be higher, correct?
 
  • #5
dgreenheck said:
Is there ever a time where you would have the same B, though?
Well, sure. I could adjust the current through both solenoids until it gave me the same magnitude of B.

There might be a limit to the magnitude of B in a solenoid if you take into account magnetic saturation, but that's beside the point.

dgreenheck said:
If the B field is always dependent on the permeability of the material it's flowing through, the squared factor of mu on top will cancel out the factor of mu in the denominator and the energy density will always be higher, correct?
Energy density just means the energy stored per unit of volume.

Consider this as a counterexample:
Assume it's possible for the sake of argument to run 1 million amps through a steel core solenoid. If I ran 1 milliamp through the same solenoid but replaced the core material with air, I'd be confident in telling you that the energy density would be higher in the case of the steel core solenoid.
 
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  • #6
I meant to say for the same current. All that is happening is the core is being inserted or removed. But I understand now, thank you.
 
  • #7
dgreenheck said:
I meant to say for the same current.
Ah, then as rock.freak667 wrote, you have:
Uair/V = 1/2*μair*n2*I2
Usteel/V = 1/2*μsteel*n2*I2 = 1/2*100*μair*n2*I2
Usteel/V = 100*Uair/V

where n is the number of turns of the solenoid.
 

1. What is a solenoid?

A solenoid is a coil of wire that is tightly wound in a helical shape. When an electric current is passed through the wire, it creates a magnetic field. Solenoids are commonly used in electronic devices such as speakers, motors, and relays.

2. What is the magnetic energy of a solenoid with a core?

The magnetic energy of a solenoid with a core is the energy stored in the magnetic field created by the solenoid. This energy is dependent on the number of turns in the coil, the current passing through it, and the permeability of the core material. It can be calculated using the equation E = (1/2)LI^2, where L is the inductance of the solenoid and I is the current.

3. How does the magnetic energy change when a core is added to a solenoid?

Adding a core to a solenoid can significantly increase the magnetic energy. This is because the core material has a higher permeability than air, allowing for a stronger magnetic field to be created within the solenoid. The inductance of the solenoid also increases with a core, leading to a higher magnetic energy.

4. What is the difference between a solenoid with and without a core?

A solenoid with a core has a higher magnetic energy and can create a stronger magnetic field compared to a solenoid without a core. This is due to the core's ability to concentrate the magnetic field lines and increase the inductance of the solenoid. Additionally, a solenoid with a core is more efficient as it requires less current to produce the same magnetic field as a solenoid without a core.

5. How is the magnetic energy of a solenoid with a core affected by the type of core material used?

The type of core material used in a solenoid can greatly affect its magnetic energy. Materials with higher permeability, such as iron or ferrite, will result in a higher magnetic energy compared to materials with lower permeability, such as air or wood. The shape and size of the core can also impact the magnetic energy of the solenoid.

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