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Force on a Current in a Magnetic Field

  1. Mar 1, 2007 #1
    Help - Force on a Current in a Magnetic Field

    1. The problem statement, all variables and given/known data
    Magnetic and gravitational forces

    A horizontal wire is hung from the ceiling of a room by two massless strings. A uniform magnetic field is directed from the ceiling to the floor. When a current exists in the wire, the wire swings upward and makes an angle with respect to the vertical (wire is in equilibrium).


    The wire has a length of 0.20 m and a mass of 0.065 kg and carries a current of 42 A. The magnitude of the magnetic field is 0.060 T.

    (e) What is the algebraic expression for the angle in the drawing? Express your answer in terms of the current I in the wire, the length L of the wire, the magnitude B of the magnetic field, the mass m of the wire, and the magnitude g of the acceleration due to gravity.

    2. Relevant equations

    F = ILB sin(theta)

    3. The attempt at a solution

    While this is a relatively easy problem, I still can't see what mistake I am making.

    First, given that the wire is in equilibrium, [​IMG]Fx = 0 and [​IMG]Fy = 0

    The three forces acting on the wire in the question are Tension in the two
    Strings, the magnetic force from the magnetic field acting on the wire, and the force of gravity.

    Using right hand rule, the magnetic force points to the right. It has to be perpendicular to the magnetic field and current (current replacing the velocity component in the right hand rule)

    Knowing this, three equations are generated:

    2T = Tension of two strings
    the tension can be resolved into x and y components
    x-component = 2Tsin(theta)
    y-component = 2Tcos(theta)

    Magnetic Force = ILBsin(theta) going to the right

    Force of Gravity = mg going down

    using Netwon's second law,
    2Tcos(theta) - mg = 0
    2T = mg / (cos(theta)) // solving for 2T using y-components

    2Tsin(theta) - ILBsin(theta) = 0

    2Tsin(theta) = ILBsin(theta) // solving for theta
    2T = ILB
    mg / (cos(theta)) = ILB // substitution from previous equation
    mg / ILB = cos(theta)

    arccos( mg / ILB ) = theta
    This is the answer I arrived at for question (e), but according to my online problem set this answer is incorrect. If any insight could be given that would be great. Thanks!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Mar 2, 2007
  2. jcsd
  3. Mar 2, 2007 #2
    There are two different angles that you have called theta. They are not the same.
  4. Mar 2, 2007 #3
    Haha yeah, i just back from school today and figured it out during the day; one of my friends pointed it out. Silly me thinking you could cancel the sin(theta)'s. Thanks for the help.
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