Strength of magnetic field at the end of a wire wrapped nail.

[kayneblue12]

Homework Statement

You want to make an electromagnet to show your friend’s kids. You have:
-Wire 2mm thick. Wire resistance 0.6Ω total
-A nail (made of mild steel)
-A battery, ∆V = 6V, with internal resistance 0.9Ω (i.e., the battery acts like a voltage source in series with a 0.9 Ω resistor.)
-You wrap the wire around the nail as tightly as you can... figure out how closely spaced the wraps will be.
-You wrap two layers (i.e., there are 2X as many wraps)
-The nail is ferromagnetic, and increases the magnetic field you get by 100X (i.e., 100X stronger than it would be if all you had was a wire coil with no nail...)
-You connect the battery to the wire around the nail and current then flows.
Figure out how strong the magnetic field you get at the end of the nail will be.

Relevant Equations

V=IR, magnetic field of solenoid B=(u0)*N*I/L ,

V=I*R
6v=I*(0.6+0.9)ohms
I=4amp
B=100*(uo)(2N)(I)/L * 1/2 I think since the wire is double wrapped, we multiply the equation by 2, but since we are looking for the magnetic field at the end of the wire we also have to multiply the equation by 1/2
I=4A, uo= 4pi*10^-7
2N/L turns per unit length, so if we have 1 turn double wrapped, the height of that wire would be 2*10^-3m and 2N/L for the double wrapped coil would be 1000 turns/meter
if we plug it into the equation for magnetic field I would get B=100*(4pi*10^-7)(1000turns/meter)(4A) /2 = 0.25T
I'm not sure if this is correct though.

 

[Charles Link]

I didn't check the complete arithmetic, but it looks to be correct. And you do have it right that the magnetic field at the end of the long solenoid is 1/2 of what it is at the center.