Magnetic field generated by current in semicircular loop at a point on axis

  1. 1. The problem statement, all variables and given/known data
    Determine the magnetic field strength and direction at a point 'z' on the axis of the centre of a semi-circular current loop of radius R.


    2. Relevant equations
    Biot Savart Formula
    [tex]d\vec{B}=\frac{\mu_{0}Id\vec{r}\times\hat{e}}{4\pi|\vec{R}-\vec{r}|^{2}}[/tex]

    e being the unit vector from r to R
    3. The attempt at a solution
    A much simpler problem is a full current loop, because one component of the magnetic field cancels out. For this problem, you'd have to deal with the half-circle arc and the straight line base separately. I was also wondering whether its easier to calculate the z and x components of B separately as well... One component is straightforward enough... I just really don't understand where to start.
     
    Last edited: Mar 13, 2010
  2. jcsd
  3. gabbagabbahey

    gabbagabbahey 5,009
    Homework Helper
    Gold Member

    This should be a pretty straightforward application of the Biot-Savart Law. Start by finding expressions for [itex]\textbf{r}[/itex], the position vector for a general point on the semi-circular arc, and [itex]\textbf{R}[/itex] the position vector for a general point on the [itex]z[/itex]-axis....what do you get for those?...What does that make [itex]\hat{\mathbf{e}}[/itex]? What is [itex]d\textbf{r}[/itex] for a semi-circualr arc?

    To makethings easier, you will want to use cylindrical coordinates.
     
  4. So, the parametric representation of a point on the semi-circle would be (0, bcos(t), bsin(t)) where b is the radius of the semi-circle.
    The vector R is just [d, 0, 0] where d is the distance on the axis of the point
    and then the e is the unit vector from R-r
    But what's dr? And where does the switch to cylindrical coord come in?
     
  5. I think I got it. Thanks
     
  6. gabbagabbahey

    gabbagabbahey 5,009
    Homework Helper
    Gold Member

    If you'd like to post your result, we''ll be able to check it for you.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?