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Acceleration of a rod due to magnetic force

  1. May 6, 2015 #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 in the figure(Figure 1) . The bar is in a uniform, horizontal, 1.60-T magnetic field and is not attached to the wires in the circuit.

    27.74.jpg
    2. Relevant equations

    Fb=I LxB
    Fg=mg
    I=V/R
    F=ma

    3. The attempt at a solution

    Resistance of bar and 10 ohm resistor are parallel, so combining them gives a ohm resistor. That is in series with the 25 ohm resistor, so the Req of the circuit is 30 ohms.
    I=V/R=120/30=4 A
    Using Kirchoff's laws, found that current passing through bar is 2 A.
    Since current is moving clockwise, right hand rule says the force due to the magnetic field is moving the bar to the right.
    The net force is Fb-Fg. Fg is given as 2.6N, Fb is
    (2)(1.6)x(1.5)
    Since L and B vectors are perpendicular, take the scalar product.
    Fb=4.8N
    To find m,
    Fg=mg => Fg/g=m => 2.6/9.8=0.27 kg
    To find acceleration, solve F=ma for a
    a=F/m
    So
    a=(Fb-Fn)/m
    a=(4.8-2.6)/.027
    a=8.15

    I got it wrong, and it says that the correct answer is 18.1 m/s^2. Any ideas as to where I went wrong? Thanks.
     
  2. jcsd
  3. May 6, 2015 #2
    Due to an initial current, it will have an acceleration and as it moves, the net flux through the loop changes. Hence another Emf is induced on that loop. So now you know why you did not get the required answer.
     
  4. May 6, 2015 #3
    Sorry, I didn't clarify. They want to know the acceleration right after the switch is flipped. I thought that the emf is only induced as a result of changing current...if we're just looking at the instant that the switch is flipped, then there wouldn't be an induced emf yet since the bar hasn't moved and the current is still 2A...is that correct?
     
  5. May 6, 2015 #4
    So if i disregard the contribution of gravity to the net force, i just get 4.8/.027=17.7, much closer to the answer provided by masteringphyics. But i don't understand why you wouldn't consider gravity...is there an underlying assumption that i missed?
     
  6. May 6, 2015 #5
    Also, Fb amd Fg are acting in perpendicular directions, since the circuit is kept horizontal. So you have to find the resultant force using vector method. You can subtract them like you have shown in the post.
     
  7. May 6, 2015 #6

    TSny

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    Homework Helper
    Gold Member

    There are three forces acting on the rod. You need to consider the direction of these forces and the direction of the acceleration of the rod. Draw a free body diagram of the rod.

    Note, the data is given to three significant figures. So, the mass should be (2.60 N)/(9.80 m/s2) = 0.265 kg (rather than 0.27 kg).
     
  8. May 6, 2015 #7
    I hadn't even considered the normal force...looking at the force diagram, the normal force would cancel the force due to gravity, leaving the net force as only the magnetic force. Is that accurate?
     
  9. May 6, 2015 #8

    TSny

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    Homework Helper
    Gold Member

    Yes.
     
  10. May 6, 2015 #9
    Okay awesome, that makes sense to me now. Thank you!
     
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