Current required to keep an iron bar in a solenoid

Question

I found myself reading some lecture notes on magnetism and tried out the challenge question: http://spiff.rit.edu/classes/phys313/lectures/sol/sol_f01_long.html (At the bottom of the page).

What I did was first calculate the volume of the bar, then the mass and then the force required to hold it against gravity, assuming g = 10.

For this I got that the force required was: ~0.02N

Using this value I tried to calculate the required magnetic field B and got: 2 * $sqrt{μ_0 / pi}$

I then tried calculating the current required and got: 1 /100 * $sqrt{1 / ( μ_0 * pi )}$

Calculating this I get ~5A.

I found the answer and dont know what I did wrong: http://spiff.rit.edu/classes/phys313/quiz/quiz.24.html

Should I have used E$_h$ = mgh somewhere?

StressBall · Answer

Thanks to Michael Richmond.

He setup the question from the link above and responded to my email query with the following:

$$n = N/l = 200/0.2 = 1000 turns / m$$

$$µ_{m} = 2times10^{-2} T/A^2$$
$$µ_{0} = 1.26times10^{-6}N/A^2$$
$$ρ = 7800 kg/m^3$$

Mass of the bar:
$$m = ρ pi r^2l = 7800(pi(0.02m)^2(0.2m)) = 1.96kg$$

Force of gravity:
$$F_{g} = mg = 19.2N$$

Magnetic force (From an energy perspective):
$$F_{B} = frac{pi r^2B^2}{2µ_{0}}(µ_{m}/µ_{0} -1) = frac{pi r^2(µ_{0}nI)^2}{2µ_{0}}(µ_{m}/µ_{0} -1) = frac{pi (0.02m)^2(µ_{0}1000I)^2}{2µ_{0}}(µ_{m}/µ_{0} -1) = 7.9times10^{-4} (15.872) I^2 = 12.6I^2$$

Therefore the current is:
$$F_{B} = F_{g} to I^2 = frac{F_{B}}{12.6} to I = 1.2A$$

Looking at this I made two mistakes;

I tried working in cubic meters and then switched to meters, getting a force of 0.02N
I accidentally simplified $$(µ_{m}/µ_{0} - 1) = 1$$

Current required to keep an iron bar in a solenoid

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