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Homework Help: A steel rod of 0.10kg rests on two metal rails inclined at an angle of

  1. Jan 15, 2012 #1
    A steel rod of 0.10kg rests on two metal rails inclined at an angle of 30 degrees. There is a magnetic field of 0.50T on the incline and a current of 70A flowing across the bar. Find the acceleration of the bar.
    The bar is 0.12m wide**

    The attempt at a solution

    What I have so far is a diagram of the problem, but I'm not sure if that is correct.
    My intuition is that you'd need to make an fbd of all the forces. On my fbd I have:
    Fn, Fg and Fb (force due to the magnetic field). I know only a portion of the 0.50T interacts
    with the rod. I'm just confused where I put the Fb in my sum of forces statements. I have two,
    Fnet (paralell) and Fnet(perpendicular) [to the motion]. Thanks!
    Last edited: Jan 15, 2012
  2. jcsd
  3. Jan 15, 2012 #2
    Re: Electromagnetism

    anyone ?
  4. Jan 16, 2012 #3
    Re: Electromagnetism

    I am not certain of the physical arrangement without seeing a diagram but the main equation you need gives the force on a conductor carrying a current in a magnetic field
    Do you know this equation
  5. Jan 16, 2012 #4
    Re: Electromagnetism

    well, what I know is the force of a wire interacting with an external B field.

    Fb = BIL
  6. Jan 16, 2012 #5
    Re: Electromagnetism

    That is correct for a field at 90 to the rod. The 30 degrees mentioned in your question must have some relevance. Do you have a diagram
    What value of force do you get using the numbers in the question?
  7. Jan 16, 2012 #6
    Re: Electromagnetism

    Well this is what I get:

    Fb = BIL(perpendicular)
    Fb = 0.5T(70A)(0.12mcos30)
    Fb = 3.6373N into the page

    Fnet (perpendicular) = Fn - mgcos30

    Fn = mgcos30
    Fn = (0.1kg)(9.8 N/kg)cos30
    Fn = 0.8487N

    There's also mgsin30, but I don't know where that goes. This is where I'm stuck.
  8. Jan 16, 2012 #7
    Re: Electromagnetism

    I still can't picture the arrangement.....if the 2 rails are at 30 to the horizontal and the magnetic field causes a force up the slope of BIL = 4.2N then the resultant force up the slope = 4.2 - (0.1 x 9.81 x Sin30) = 4.2 - 0.49 = 3.71N
    does this fit with your arrangement?
  9. Jan 17, 2012 #8
    Re: Electromagnetism

    I think I've figured it out, It's just like a box on an incline question, however the current is going out of page, resulting in a magnetic force to the right up the ramp.
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