Magnetic field outside a solenoid crossed by a current.

[maCrobo]

Homework Statement

A student makes an electromagnet by winding 320 turns of wire around a wooden cylinder of diameter 4,80 cm. The coil is connected to a battery producing a current of 4.20 A in the wire. At what axial distance z>>d will the magnetic field of this dipole be 5.0 μT?

Homework Equations

Biot-Savard Law: (in vector form) B=(μ/4π) i(L x u)/r^2
Maybe Ampère's Law: ∫Bds=μi

The Attempt at a Solution

Unfortunately I can't even imagine how to integrate...
I tried by using Ampère's Law on a circular path around the solenoid. I mean: imagine the solenoid and an imaginary circumference that lays on the same plane of the coils, then inside this circumference there's the current i. I actually treated it as a straight wire because the result is the same and it's wrong...

mhh, help?

 

[maCrobo]

Well... I did it...
I got confused by "the axial distance", it meant measuring the distance along the z axis starting from the center of the coil (solenoid).
Let's start from the general formula for a circular wire carrying a current i, [itex]B=\frac{µ_{0}iR^{2}}{2\left( R^{2}+z^{2} \right)^{\frac{3}{2}}}[/itex]. Then, we approximate this formula to the case z>>R, so that [itex]B=\frac{µ_{0}iR^{2}}{2z^{3}}[/itex]. This is the result for a singular coil, our solenoid is made by 320 coils whose magnetic dipole we can know. So, I get to [itex]B=\frac{µ_{0}i\pi R^{2}}{2\pi z^{3}}=\frac{µ_{0}i}{2\pi }\frac{Nµ}{z^{3}}[/itex]. From here it's easy to get z because we know B.