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Gre Problem #87

  1. Sep 2, 2004 #1
    I am so glad I found the link to this test, so I can reference questions with figures.
    let me write the problem first, and provide the link at the end, and in between explain where I was stuck.

    87) two small pith balls, each carrying a charge q, are attached to the ends of a light rod of length d, which is suspended from the ceiling by a thin torsion-free fiber, as shown in the figure above ( in the link). There is a uniform magnetic field B, pointing straight down, in the cylindrical region of radius R around the fiber. The system is initially at rest. If the magnetic field is turned off, which of the following describes what happens to the system?

    I found the emf induced around the loop of Radius R, and I integrated from 0 to R to find the torque on the bar, and I multiplied my answer by 2, because there are two fields on the bar of opposite direction ( a loop), which will add to the torque, so I got that the bar will rotate with angular momentum = qBR^2/2,
    but the correct answer is twice as much.

    here is it with the figure:
    http://phys.columbia.edu/~hbar/Physics-GRE.pdf [Broken]

    thankyou so much!
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Sep 5, 2004 #2


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    I do not understand what and why did you integrate to get the torque. The electric force acting on one of the charges is F=qE. The torque is this force multiplied by d/2, and you have to take it twice because of the two charges.
    Here is my solution:

    I assume a uniform decrease of the magnetic field from B to 0 in [tex]\Delta T[/tex] time. The actual time dependence does not really matter. The flux inside the circle of radius d/2 is then

    [tex] \Phi = r^2\pi B(1-t/\Delta T) [/tex]
    I use the integral form of Maxwell's equation for the circle of diameter d.
    [tex] \oint {E_s ds} =-{\frac{d\Phi}{dt}}[/tex]
    As the electric field is uniform along the circle and its direction is tangent of it
    [tex]E_s = \frac{Br^2}{d\Delta T}[/tex]

    The force on one charge is [tex]F=qE_s[/tex] and the torque with respect to the centre of the circle is [tex]Fd/2[/tex]. There are two charges, so he resultant torque is [tex]M=Fd[/tex].
    [tex] M = Fd =qd \frac{Br^2}{d\Delta T} = q \frac{Br^2}{\Delta T} [/tex].
    The torque equals the time derivative of the angular momentum [tex]\Delta L/\Delta T=M[/tex]. In time [tex]\Delta T [/tex], the angular momemtum raises from zero to the value [tex]L=qBr^2 [/tex].

  4. Sep 5, 2004 #3


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    I agree with the other poster. His solution is correct.

    It sounds as if you might have picked up an extra factor of 2 by integrating r dr. But as the other poster pointed out, you should not integrate because the charges are localized at the extremities of the bar. What you did would be correct if the charges were uniformly distributed along the length of the bar, in which case the total force would be smaller. But in this problem, the charges are at the extremities and there is no integral left to do (other than the trivial [itex] \int \vec E \cdot \vec ds = 2 \pi d/2 E =\pi d \times E [/itex].

    Last edited by a moderator: May 1, 2017
  5. Sep 6, 2004 #4
    Hello ehid and pat,
    the reason I integrated is that I thought that the tube is going to suck up the charges, so they will be distributed over the whole tube ( but it is a rod, so it might be an insulator), but now that I see your solutions, it makes more sense not to think that way, and I also thought that there wouldn't be an electric field where there is no magnetic field, I was thinking narrowly, I thought that the electric field is going to show up only in the area where there was a magnetic field, but now I can see that I was wrong, I am so glad I found this forum! :smile: Many thanks for you guys!
    Last edited: Sep 6, 2004
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