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What is the magnitude of force experienced by a cylindrical piece of iron placed near the end of a solenoid with internal field strength B

_{0}.

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What is the magnitude of force experienced by a cylindrical piece of iron placed near the end of a solenoid with internal field strength B

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Vanadium 50

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Ok, then how do i calculate the force if I know the gradient of the field i.e. I know the function B(x,y,z).It's dependent on the gradient of the field strength.

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One way to calculate the force is to examine how the total stored magnetic energy W of the system changes if the iron object is displaced by a distance dx. As you know, work (energy) is force F times distance dx. So the derivative of the stored energy W in the direction dx is the force in the direction dx:Ok, then how do i calculate the force if I know the gradient of the field i.e. I know the function B(x,y,z).

F

Smythe's (Static and Dynamic Electricity) exact solution for the axial field of a finite length solenoid is given in post #4 (see thumbnail) of this thread:

https://www.physicsforums.com/showthread.php?t=352947

Bob S

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Thanks for that. It really helped me.

So can I proceed like this to calculate the force on a small iron piece lying some distance from the axis of the solenoid.?One way to calculate the force is to examine how the total stored magnetic energy W of the system changes if the iron object is displaced by a distance dx. As you know, work (energy) is force F times distance dx. So the derivative of the stored energy W in the direction dx is the force in the direction dx:

F_{x}= dW/dx.

1. The B inside the iron is multiplied by factor u

2.The total Energy stored inside the piece can be calculated by intergrating over the entire volume of the iron piece

http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/imgele/engbf.gif [Broken]

3. When the iron piece is moved a small distance then then the total Energy Stored will change.

Then applying F

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- #6

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(1/2μ

and inside the iron it is

(1/2μμ

so because B is continuous there is very little energy stored inside the iron.

But there is a dipole field outside the iron that I have not included.

I know there is a thread on this but I cannot find it.

Bob S

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on the contrarary I think there is lot of energy inside the iron becauseInside the iron it is

(1/2μμ_{0})∫B^{2}dV

so because B is continuous there is very little energy stored inside the iron.

the total energy inside iron, as you pointed out, is

(1/2μμ

But, B = μB

So, the energy stored becomes

(1/2μμ

= (1/2μμ

= (μ/2μ

When the piece of iron is moved a small distance towards the solenoid, the intensity of B increases, so does the energy stored inside the iron.

So, the iron shouldn't move towards the solenoid (since it creates increased energy).

This is directly against the fact that the iron piece is attracted towards the solenoid!!!??

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- #8

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Err..... no. Very little energy is stored in the iron. Because the normal component of B is continuous, the energy density is higher in the air than in the iron. See thumbnail.on the contrarary I think there is lot of energy inside the iron because

the total energy inside iron, as you pointed out, is

(1/2μμ_{0})∫B^{2}dV

But, B = μB_{0}

So, the energy stored becomes

(1/2μμ_{0})∫(Bμ)^{2}dV

= (1/2μμ_{0})μ^{2}∫B^{2}dV

= (μ/2μ_{0})∫B^{2}dV

When the piece of iron is moved a small distance towards the solenoid, the intensity of B increases, so does the energy stored inside the iron.

So, the iron shouldn't move towards the solenoid (since it creates increased energy).

This is directly against the fact that the iron piece is attracted towards the solenoid!!!??

Bob S

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- #9

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But In wikipaedia I read this:Err..... no. Very little energy is stored in the iron. Because the normal component of B is continuous, the energy density is higher in the air than in the iron.

Bob S

(here u is the total energy stored).

I am confused that, Is it B or H , that is same for both air core and iron-core solenoid?

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W = (1/2)∫ B H dV

In a magnetic system that has both iron and air in the same flux, only B (the normal component) is continuous at the air / iron interface. So we write

W = (1/2μ

W = (1/2μ

Bob S

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