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Homework Help: Time-Varying Magnetic Field

  1. Oct 19, 2006 #1

    The link below gives the question I'm interested in. The question is p6.2.


    My Thoughts

    OK, now I thought I knew how to do this question. The expression for the magnetic field shows that it varies with the position along y and varies with time.

    So what I thought you'd do is this:

    [tex]\Phi = \int{B.ds}[/tex]

    Where the differential area is given as:

    [tex]dS_z=dx dy[/tex]

    So the above integral becomes:

    [tex]\Phi = b B_0 \int{B.dy}[/tex]

    where b is dx (0.1) and [tex]B_0=3 \mu T[/tex] and B is the rest of the expression for the varying magnetic field.

    So I thought you'd integrate this expression with limits of 0 to 0.3, but the answer I get is nothing like their answer??

    I end up with an expression for the current i of:

    [tex]i=- \frac {B_0 b \omega} {R k}[ cos(\omega t -k a)-cos(\omega t)][/tex]

    This is nothing like their expression of:

    [tex]i=- \frac {B_0 b \omega} {R k} [sin(\frac {1} {2} k a) sin(\omega t- \frac {1} {2} k a)][/tex]

    This is simplified by them in the hints section (just scroll down the page to find the hints).

    So what am I doing wrong here??

    EDIT: oops my expressions for I are meant to contain cosines and not sines.
    Last edited by a moderator: Apr 22, 2017
  2. jcsd
  3. Oct 19, 2006 #2
    Try using this obscure trig property:

    [tex] sin(x) - sin(y) = 2 sin(\frac{x-y}{2})cos(\frac{x+y}{2}) [/tex]

    See if that helps, or throw some numbers in and graph to see if they are the same.
  4. Oct 19, 2006 #3
    Thank you so much Norman!!! That works out perfectly in getting the expression that they have.

    Took me a while to find the trigonometric product formulae.

    This probably was annoying the crap out of me because I was sure I had it right and it's just such a relief to have it fully completed.

    Thanks again!
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