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!

Rotating Bar in uniform B Field

  1. Nov 8, 2014 #1
    1. The problem statement, all variables and given/known data
    A 1.60 m long rod rotates about an axis through one end and perpendicular to the rod, with a rotational frequency char21.png of 6.91 radians per second. The plane of rotation of the rod is perpendicular to a uniform magnetic field of 0.30 T. Calculate the magnitude of the char0F.png mf induced between the ends of the rod.

    2. Relevant equations
    -dΦ/dt = ε
    sin(kx-ωt)

    3. The attempt at a solution
    Well, I'm completely lost. I think part of it is I'm not quite sure the picture they're trying to paint (is it just me or does the wording sound like garbage? I copy and pasted too), but also, I'm not quite sure how I can relate ω with ε. I think the relevant equations are how to go about it . . . but I'm not 100% sure. Do I just integrate the sin function with respect to t? But there's not enough information to do that (I don't think)?

    Also, thank you Physics Forum for helping me ace my last exam. I didn't fail (y)
     
  2. jcsd
  3. Nov 8, 2014 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    In what way do you suspect that is relevant? Is there anything, apart from the angle of the rod, that is changing over time?

    The description makes sense to me. In Cartesian terms, the rod is rotating in the XY plane, with one end fixed at the origin. The magnetic field is constant, perpendicular to the XY plane.
    Consider a short section of the rod, length dr, distance r from the origin. How fast is it moving across the field?
     
  4. Nov 8, 2014 #3
    . . . It had ω :s :L

    So if ω = 6.91 rad/sec, and total distance is 2πr . . . ωr = v?
     
  5. Nov 8, 2014 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes. So what potential difference is generated along the short section?
     
  6. Nov 8, 2014 #5
    Uh . . . I think it's a derivative but I don't have the exact relationships in front of me . . . make me go 'duh!'
     
  7. Nov 8, 2014 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

  8. Nov 9, 2014 #7

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Consider the "Blv" law: d(emf) = (B x dl) * v. Then integrate. (The asterisk is supposed to be a dot).
    Are we 'duh'ing yet? :smile:
     
  9. Nov 10, 2014 #8
    Haha, I am 'duh'ing a little bit ;) This is the answer I found, but want to type it out to make sure I'm understanding the physics of it.

    ε = (d/dt)(∫B•dA). It's a law.

    Now the solution uses area as πr^2, so I'm assuming they're just taking the area swept out by the rod to get that. Now this is where I get lost (and I'm applying an equation I found from another, similar problem). The equation, the integrand, is B•πr^2•([ωt]/[2π]).

    Where does the ([ωt]/[2π]) term come from? The rest makes sense: just deriving it. Is that the velocity you're talking about? ([ωt]/[2π])?
     
  10. Nov 10, 2014 #9

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Your emf expression is right. Like you said, it's a law!
    The formula you dug up is the flux swept by the rod in time t if the rod has length r.

    In any case it's much more edifying if you solve the problem on basic principles.
    There are 2 ways of solving this:
    1. emf ← Blv law. Integration required.
    2. emf = -dφ/dt. No integration required. This is your 'law'. So let's go with it:
    emf = -dφ/dt
    So emf = -B dA/dt, A = area swept by the rod.
    So what is the rate of area A swept, in sq. meters per sec.?

    (The following is advanced theory: sometimes No. 2 can give a wrong result. Whenever there is moving media, as in this case, the Blv law gives the right answer while #2 might not. If B is time-changing also, you need both! Don't sweat this now.)
     
    Last edited: Nov 10, 2014
  11. Nov 10, 2014 #10
    So by number two, the [ωt]/[2π] is the rate, and the πr^2 the area?
     
  12. Nov 10, 2014 #11

    rude man

    User Avatar
    Homework Helper
    Gold Member

    No, wt/2pi * pi R^2 = area swept in time t. To get the area swept per second, let t=1 sec. So now area swept per second = rate of sweeping the area = w/2pi * pi R^2 = w/2 * R^2.
     
  13. Nov 10, 2014 #12

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi there, ignore me if you don't feel comfortable with this approach, but I want to throw in a sales pitch for doing this using the Lorentz force. Because that's a bit more fundamental than the other two.

    ##\displaystyle \vec F_L = q\;\vec v \times \vec B\ ## and ##\displaystyle qV = -\int_0^l \vec F_L \cdot d\vec r## and ##\vec v = \vec \omega \times \vec r##

    A nice opportunity to use ##\vec a\times\vec b\times \vec c= \vec b (\vec a\cdot \vec c) - \vec c(\vec a\cdot\vec b)##
     
  14. Nov 10, 2014 #13

    rude man

    User Avatar
    Homework Helper
    Gold Member

    @BvU, I'm not decompiling your latex text.

    However, the Blv law is based on the Lorentz force:
    F = qvB so E = F/q = vB
    emf = integral E*dl
    so emf = Blv
    The two ways of doing this are the only two ways I've encountered.

    Note: the basic "Blv law" is d(emf) = (B x dl) * v
    where in this problem v = v(l) = ωl so an integration over l must be carried out.
     
    Last edited: Nov 10, 2014
  15. Nov 11, 2014 #14

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think we agree completely. based on ##\equiv## Lorentz is more fundamental :)

    I had a good time deriving ##\displaystyle V = - \int_0^l (\vec \omega \times \vec r \times \vec B)\cdot d\vec r ## with the triple cross expression. But yes, it's an integral.

    In fact ##\vec v = \vec \omega \times \vec r## is a triple cross product too, since ##\vec r \times m \vec v \equiv \vec L = I\vec \omega = mr^2 \;\vec \omega\ \rightarrow \ \vec \omega \times \vec r = \hat r \times \vec v \times \hat r =\vec v\ ## because ##\vec r\cdot \vec v = 0## here.

    Enough said; back to work/business :)

    Latex makes it look cute, but I waste time to get the little arrows over ##\vec a\vec b \vec c## at the same height :(
     
    Last edited: Nov 11, 2014
  16. Nov 11, 2014 #15

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Yes, as I said, when there are moving media the Blv law is the better choice since Maxwell/Faraday can get you into trouble.

    I personally refuse to learn latex and am still very disappointed that the old symbols table has been removed. I think ehild is the only poster who graciously includes the table in all his posts.
     
  17. Nov 11, 2014 #16

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There's a lot of symbols when you click the ∑ in the toolbar...

    TeX is a time-eater I admit, but its a fun addiction :)

    How do you know ehild is a he and not a she ?
     
  18. Nov 11, 2014 #17

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Thank you for the tip!
    And you're right, I don't know ehild is a 'he'. Do you know? He/she is very knowledgeable. Are you and he/she perhaps from the same country? Like Hungary?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Rotating Bar in uniform B Field
  1. Rotating a B field (Replies: 0)

Loading...