Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Magnetic field outside a solenoid crossed by a current.

  1. Jan 6, 2012 #1
    1. The problem statement, all variables and given/known data
    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?

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

    3. 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?
  2. jcsd
  3. Jan 7, 2012 #2
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook