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Can not understand how to divide the components of the vector.

  1. Jan 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a closed semi-circular loop lying in the xy plane carrying a current I in the
    counterclockwise direction, as shown in Figure attached.

    2.png

    A uniform magnetic field pointing in the +y direction is applied. Find the magnetic force acting on the straight segment and the semicircular arc.



    2. Relevant equations

    The force in xx axis [itex]\vec{F1}[/itex] is easy to see that it has the normal direction [itex]\hat{i}[/itex]

    so that force :

    [itex] \vec{F} = I 2R\hat{i}\times \vec{B} \hat{j}= 2IRB \hat{k} [/itex]

    where [itex]\hat{k}[/itex] is directed out the page.

    Now the force along the arc.


    The solution says:

    To evaluate [itex]\vec{F2}[/itex] , we first note that the differential length element [itex]d\vec{s}[/itex] on the semicircle can be written as:

    [itex]d\vec{s} = ds\hat{\theta } = IRd\theta (-sin \theta \hat{i} + cos \theta \hat{j})[/itex]


    I know that [itex]s = R \theta[/itex], but i don't know where [itex]-sin \theta \hat{i} + cos \theta \hat{j} [/itex] come from.



    Some tips ??
     
  2. jcsd
  3. Jan 5, 2013 #2

    TSny

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    Gold Member

    The vector ##\vec{ds}## at a point on the circular arc is tangent to the circular arc and therefore perpendicular to the radius. (The diagram doesn't really show it as perpendicular, but that's because they've drawn a fairly large ##\vec{ds}## vector. It is actually infinitesimally small.) ##\theta## is the angle that the radius line makes to the horizontal. Can you deduce the angle that ##\vec{ds}## makes to the horizontal or the vertical?

    Should your expression [itex]d\vec{s} = ds\hat{\theta } = IRd\theta (-sin \theta \hat{i} + cos \theta \hat{j})[/itex] have the current I in it?
     
  4. Jan 5, 2013 #3
    Actually it hasn't the current in the expression (my mistake)

    At first glance it seems that [itex]d\vec{s}[/itex] makes a [itex]\theta [/itex] angle with the horizontal. But that doesn't make sense because the component of the horizontal would be [itex](-cos \theta \hat{i}[/itex])
     
    Last edited: Jan 5, 2013
  5. Jan 5, 2013 #4
    [itex]-sin \theta \hat{i} + cos \theta \hat{j} [/itex] is a unit vector pointing in the θ direction, expressed in terms of the unit vectors for the x-y cartesian coordinate system. The differential position vector ds is pointing in the θ direction. Its magnitude is Rdθ.
     
  6. Jan 5, 2013 #5
    But the problem is that i don't understand why it is not [itex]-cos \theta \hat{i} + sin \theta \hat{j}[/itex] because the vector [itex]\vec{ds}[/itex] makes a angle [itex]theta[/itex] with the horizontal.
     
  7. Jan 5, 2013 #6
    hum, maybe now it is correct?

    p5.png

    The angle that is does with the horizontal is [itex]90 - \theta[/itex] so it is [itex]cos(90 - \theta) = sin(\theta) [/itex]
     
  8. Jan 5, 2013 #7

    TSny

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    Yes, that's right.
     
  9. Jan 5, 2013 #8
    Now it is understood!

    Thanks again!
     
  10. Jan 5, 2013 #9
    A trick that helps sometimes (or at least allows you to check your answer) is to consider the unit vector at the limits θ=0 and at θ=π/2. If you are considering the unit vector in the θ direction, then at θ=0, the θ-direction unit vector is +j, and at θ=π/2, the θ-direction unit vector is -i. We also know that at θ=0, cosθ=1, and sinθ=0, while at θ=π/2, cosθ=0, and sinθ=1. Therefore, the coefficient of j must be cosθ, and the coefficient of i must be -sinθ.
     
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