1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Magnetic Fields and Equilibrium

  1. Apr 16, 2007 #1
    1. The problem statement, all variables and given/known data

    A wire is bent into a square connected to the terminals of a 9 Volt battery. The wire is constructed of copper with diameter of 1 mm and is bent into a square with side length of 30 cm. If the battery has a mass of 50 g and
    internal resistance 0.1 Ω, what magnetic field (magnitude and direction) is needed in the shaded region to keep the system in equilibrium? The resistivity of copper is 1.7 x 10^(-8) Ω m with a density of 8920 kg/cubic meter.

    [​IMG]

    2. Relevant equations

    B = (μI)/(2πL)
    Force in a wire = LI x B = LIB
    V = IR
    r = internal resistance = 0.1 Ω
    Resistivity ρ = (RA)/L
    d = 0.3 m

    3. The attempt at a solution

    Well...I know that net force = 0 if the system is in equilibrium. The forces that I have identified are the weight and the magnetic force. Should I consider electric force as a possible force? Why do I even need the mass of the battery?

    Resistivity ρ = (RA/L)
    r = (ρL)/A = (1.7 x 10^(-8) Ω m)(0.3 m)/(π x 0.0005² m²) = 0.0065 Ω
    I = V/R = 9V/(0.1 Ω + 0.0065 Ω) = 84.5 A
    current flows clockwise from the positive terminal to the negative terminal.
    B = (μI)/(2πL) = [(4π x 10^-7 Tm/A)(84.5 A)]/(2π x 0.3 m) = 5.63 x 10^-5 T
    length L = 0.3 m
    μ = 4π x 10^-7 Tm/A
    Mass of the top part of the wire = (density)(volume) = (8920 kg/cubic meter)(π x 0.0005² m²)(0.3 m) = 0.0021 kg
    weight = mg = (0.0021 kg)
    Net force = magnetic force - weight = 0
    LIB - mg = 0
    LIB = mg
    B = (mg)/(LI) = (0.0021 kg x 9.8 m/s²)/(0.3 m x 84.5 A) = 8.12 x 10^-4 T

    In order to find the direction of the magnetic field, I tried using the right hand rule and found that the magnetic field points into the page. Does this seem to make any sense? I also think that the two parallel current carrying wires in the shaded region have currents in opposite directions so their forces should also cause them to repel.

    After solving it two different ways, I got two different answers of B = 8.12 x 10^-4 T and B = 5.63 x 10^-5 T. I'm also having trouble telling if the magnetic field points inward or outward; I think it points inward after using the right hand rule. Please help if you can. Thank you so much.
     
  2. jcsd
  3. Apr 16, 2007 #2

    hage567

    User Avatar
    Homework Helper

    One thing I see is that when you are trying to find the current in the wire, you should take L to be the length of the entire loop, not just the one side. So it should be 1.2 m instead of 0.3 m. The same thing goes for finding the total mass of the wire.

    I don't think B = (μI)/(2πL) applies to this problem, as you are trying to find the magnitude of an external magnetic field that will provide the upwards force. This equation is for the magnetic field due to current in a wire.

    If the current is clockwise, then the magnetic field will point into the page for an upwards force.
     
  4. Apr 16, 2007 #3
    Thanks for your help, so now I get...
    Net force = Magnetic Force - weight = 0
    LIB - mg = 0
    LIB = mg
    Mass of the wire = (density)(volume) = (8920 kg/cubic meter)(π x 0.0005² m²)(1.2 m) = 0.0084 kg
    B = (mg)/(LI) = (0.0084 kg x 9.8 m/s²)/(1.2 m x 84.5 A) = 8.125 x 10^-4 T

    Does this look better? Thank you so much for your help. I really appreciate it, but I still don't understand why the problem states the mass of the battery.
     
  5. Apr 16, 2007 #4

    hage567

    User Avatar
    Homework Helper

    The mass of the battery is needed because it is providing a downwards force, so it must be included so the magnetic field is strong enough to hold it up (I am assuming the battery is not touching the ground, that is the point of the problem).

    I don't think your current value is right (it didn't appear to change anyway), you must include the entire length of the wire when you are calculating the resistance of the wire (since the current goes through the entire wire).

    Now, when you get to this last step of calculating B, you want to use only 0.3 m for L, since it is only the top section of the wire that is interacting with the magnetic field (the vertical parts of the loop don't contribute to the upwards force).

    Your approach looks OK you just have to work out a few of the little details. Go through it very carefully and check that everything makes sense!
     
  6. Apr 16, 2007 #5
    Oh, ok. That makes sense. Wow, I am so incompetent.
    Resistivity ρ = (rA/L)
    r = (ρL)/A = (1.7 x 10^(-8) Ω m)(1.2 m)/(π x 0.0005² m²) = 0.0260 Ω
    I = V/R = 9V/(0.1 Ω + 0.0260 Ω) = 71.4 A
    Net force = Magnetic Force - weight of wire - weight of battery = 0
    Magnetic Force = weight of wire + weight of battery
    LIB = (mass of wire)g + (mass of battery)g
    Mass of the wire = (density)(volume) = (8920 kg/cubic meter)(π x 0.0005² m²)(1.2 m) = 0.0084 kg
    B = (mg + mg)/(LI) = (0.0084 kg x 9.8 m/s² + 0.05 kg x 9.8 m/s²)/(0.3 m x 71.4 A) = 0.0267 T
    The magnetic force points upward so the magnetic field must point inward via the right hand rule.

    Wow, my answer is now totally different from my previous answer. I hope this is right. Thank you so much for all your help! :)
     
  7. Apr 16, 2007 #6

    hage567

    User Avatar
    Homework Helper

    You're welcome. I hope it's right too! I think it should be.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Magnetic Fields and Equilibrium
  1. Magnetic Field (Replies: 2)

  2. Magnetic field (Replies: 2)

  3. Magnetic fields (Replies: 3)

  4. Magnetic Field (Replies: 2)

Loading...