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Homework Help: Help with B field for a homemade electromagnet

  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm doing a physics demo, and I've made basic electromagnet using some copper wire, a 6 volt battery, and a carbon steel screw, and I need help checking my calculations because I think the magnetic field for my magnet is too large. I've double checked everything and can't come up with why it's so large....(The reason I think it's not nearly as strong is I have a a neodymium magnet to compare it with, which has ~ 1.25 T B field, and although it is close to it's strength, it is not as strong). The only thing that makes the B-field believable to me in terms of strength is removing the relative permeability constant.

    Variables used in my calculations:
    Wire characteristics: 7.3152 m, 16 AWG (cross sectional area of 1.330763e-6 m2)
    Solenoid Core: .135m long carbon steel screw with a radius of approx 0.05m
    6 volt battery
    μ0 = 4π × 10−7 H/m
    μcarbon steel1.26×10−4 H/m
    2. Relevant equations


    I = V/R


    3. The attempt at a solution

    Wire: 24 ft. (7.3152 m), 16 AWG
    Solenoid Core: 0.135m long Carbon Steel Screw with a radius of approx. 0.05m

    Carbon Steel magnetic permeability (μ) = 1.26e-4 H/m => a relative permeability of ~100 (μ/μ0)

    Resistance = ρl/A

    ρ = 1.724e-8 ohm*m, l = 7.3152 m, A= 1.330763e-6 m2

    Resistance = 0.0964 ohms

    I = V/R

    V = 6 V, R = 0.0964 ohms => I= 62.2118 Amperes

    B=kμ0nI μ0 = 4πe-7 T*m/A n= 143 turns/0.135 m = 1059.259259 turns/m k=relative permeability of carbon steel = 100

    B = (100)(4πe-7 T*m/A)( 1059.259259 turns/m)( 62.2118 Amperes) = 8.281 Tesla
    Last edited: Dec 10, 2014
  2. jcsd
  3. Dec 10, 2014 #2

    rude man

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    OK to check: (I like using B = μNi/L ~ 1.3e-4 H/m * 143 turns * 62.2A / 0.135m ~ 8.6T.
    But here's the thing: what kind of battery are you running? The power dissipation in your wire is a whopping 348W which, to be frank, I wonder about. Did you measure your battery voltage when the wire was connected?
    Otherwise your calculations look good.
  4. Dec 10, 2014 #3
    I'm using a Eveready 6 Volt Lantern Battery 1209, and unfortunately I don't have a voltmeter to test it at the moment. So, no, I didn't measure it when the wire was connected.
  5. Dec 10, 2014 #4
    Okay, I found an old multimeter, and I belive I've got it measuring correctly, reading almost exactly 6 volts without the wire, and ~.5 volts with the wires hooked up.
  6. Dec 10, 2014 #5

    rude man

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    That's more like it. So compute your current as 0.5V/0.09ohms= 5.6A. The coil should still feel hot at ~ 2.8W dissipation. And the 0.5V might not hold up for long so monitor that too.
  7. Dec 10, 2014 #6
    Awesome, thank you, that makes everything seem way more believable. I appreciate the help greatly.
  8. Dec 11, 2014 #7

    rude man

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    You're most welcome.
  9. Dec 11, 2016 #8
    62.2118 Amperes is impossible to reach from a 6 volt battery because of its internal resistance (6 volt battery short circuit current is <20 amperes approx.)
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