# Magnetic Energy of Solenoid With/Without Core

1. May 20, 2013

### dgreenheck

I have a question about solenoids.

The formula for the magnetic field energy density is:

$\frac{1}{2}\frac{B^{2}}{μ}$

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?

2. May 20, 2013

### rock.freak667

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. May 20, 2013

### milesyoung

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. May 22, 2013

### dgreenheck

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. May 22, 2013

### milesyoung

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.

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.

Last edited: May 22, 2013
6. May 22, 2013

### dgreenheck

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. May 22, 2013

### milesyoung

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.