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Acceleration of Bar in Circuit with Magnetic Field

  1. Mar 30, 2014 #1
    1. The problem statement, all variables and given/known data
    A 2.60-N metal bar, 1.50 m long and having a resistance of 10.0 Ω, rests horizontally on conducting wires connecting it to the circuit shown below. The bar is in a uniform, horizontal, 1.60-T magnetic field and is not attached to the wires in the circuit. What is the acceleration of the bar just after the switch S is closed?

    OI9ZbAl.jpg

    2. Relevant equations
    Well, the magnetic force of a straight wire is valid here.
    F = IlB‎sin‎Ø

    And Newton's Second Law as well. Fnet = ma.
    So combining the two equations, I get IlBsin‎Ø - mg = ma and thus a = (IlBsin‎Ø - mg)/m.

    Also, since we have the potential difference of the source, ε = IR is valid as well.

    3. The attempt at a solution
    The equivalent of the parallel resistors in 5.0 Ω and the total (equivalent of the series) is 30.0 Ω.
    From that,
    ε = IR
    I = ε/R
    I = (120 V)/(30 Ω)
    I = 4.0 A

    And the path splits in two before it reaches the bar, so the current that will pass through the bar is 2.0 A.

    However, the application of N2L is confusing me. We are not given either the mass of the bar or the acceleration; just the net force of the bar (2.60-N). How do I determine the mass to use in N2L and ultimately obtain the acceleration?

    Thanks.
     
  2. jcsd
  3. Mar 30, 2014 #2
    I'm going to assume the 2.60 N is the weight of the bar. Knowing the gravitational acceleration of whatever planet the circuit is on (surface of Earth is probably a safe bet) should give you the mass of the bar.
     
  4. Mar 30, 2014 #3
    That makes sense. However, it doesn't explain why the solutions manual does not consider the weight force of the bar when applying's N2L:

    They solve
    a = IlB/m
    where m = 2.60 N/9.80 m/s^2 as you said. But I don't understand why it is not a = (IlB - mg)/m.
     
  5. Mar 30, 2014 #4
    It's implied that the bar is restricted to horizontal movement. Gravity has no horizontal component, so it doesn't contribute to the net force acting on the bar in the horizontal direction.
     
  6. Mar 30, 2014 #5
    That makes perfect sense. Thank you so much.
     
  7. Mar 30, 2014 #6
    You're welcome. Consider also that in Newton's second law:
    [tex]
    \mathbf{F} = m \mathbf{a} \quad (1)
    [/tex]
    ##\mathbf{F}## and ##\mathbf{a}## are vectors, but we can express (1) component-wise as:
    [tex]
    \begin{align}
    F_x &= m a_x \quad (2)\\
    F_y &= m a_y \quad (3)
    \end{align}
    [/tex]
    where x and y subscripts refer to horizontal and vertical components, respectively.

    In your example, we know that ##a_y## is zero due to physical restrictions, so we're really only concerned with (2). I hope you can see why:
    mixes up (2) and (3).
     
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