1. Limited time only! Sign up for a free 30min personal 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!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A steel rod of 0.10kg rests on two metal rails inclined at an angle of
Loading...