So I still not sure how to apply like Right-hand rule (RHR) in this setup in problem like the one in the following so I tried to do RHR in order to get the direction but it didn't work out. This is an example from the Halliday and Resnick book.

Figure 32-24 shows a wire segment, placed in a uniform magnetic field B that points out of the plane of the figure. If the segment carries a current $i$, what resultant magnetic force acts on it ?

Here is image of the problem at hand; $F_1$ and $F_3$ are the force on the straight wire segments enter image description here


$$F_1 = F_3 = -iLB $$

so its equal to

$$iL = \langle iL,0,0\rangle \quad B = \langle0,0,B\rangle$$

$$iL\times B = \langle0,-iLB,0\rangle$$ So it points in the negative y direction I tried to do same thing now for circular part to get the directions:

$$iL = i\langle r \cos \theta,r \sin \theta ,0\rangle \quad B =\langle0,0,B\rangle$$

$$iL \times B = \langle r \sin\theta B,r \cos\theta,0\rangle$$

So since its a circle, $\cos \theta$ will gets canceled by symmetry so it will be so it will have only a horizontal component equal to $r \sin \theta B$ of course this works out in the problem when you do the integral but its not right reasoning as the book have and book used RHR.

  • 3
    $\begingroup$ Hi user34615, please try to put more effort into your questions. Without spaces in your formulas your question is almost incomprehensible. The TeX markup feature is there for a reason. $\endgroup$ – Brandon Enright Nov 24 '13 at 18:08
  • $\begingroup$ alright I will next time its first time using physics stack exchange. $\endgroup$ – Illustionisttt. Nov 24 '13 at 18:20
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    $\begingroup$ The edit button is there for a reason too. $\endgroup$ – Brandon Enright Nov 24 '13 at 18:20

There are a lot of details missing in your question but I will try to get you in the right direction.

First, whenever you write vector components, you need to be clear in what coordinate system you're using. I will assume that $\mathrm{OX}$ points to the right in your figure, $\mathrm{OY}$ points up, and $\mathrm{OZ}$ points out of the screen (so that it is right-handed). For the sake of clarity, the unit vectors pointing in their corresponding directions are $\hat{\imath}$, $\hat{\jmath}$ and $\hat{k}$ respectively. In this coordinate system, your magnetic field would be written as $\vec{B}=B\hat{k}$.

Second, you didn't specify the direction in which the current $I$ is travelling. I will further assume it travels in the direction of $\hat{\imath}$ in the straight segments and, therefore, clockwise in the arc segment.

By using $\mathrm{d}\vec{F}=I\,\mathrm{d}\vec{L}\times\vec{B}$ you can figure the resulting force on each segment. I will focus in the right-hand rule to figure the direction of the resulting force.

Straight segments: since $\mathrm{d}\vec{L}$ points along $\hat{\imath}$ in every differential of these segments, and the magnetic field points along $\hat{k}$, then you need to point your right hand's four fingers in the direction of $\vec{L}$ ($\hat{\imath}$ for these segments) so that you can close your hand to the direction of $\hat{B}$ $\hat{k}$ in this problem). By doing so, your thumb should end up pointing along $-\hat{\jmath}$. By noting $\vec{L}$ and $\vec{B}$ are perpendicular, you can write $\vec{F}_1=\vec{F}_3 = -ILB\hat{\jmath}$.

Arc: at each differential of the wire $\mathrm{d}\vec{L}$ is tangent to the wire. More specifically, $$\mathrm{d}\vec{L} = R(\sin\theta\hat\imath-\cos\theta\hat\jmath)\mathrm{d}\theta$$

If you use the right hand rule as described above (don't forget $\mathrm{d}\vec{L}$ changes direction with $\theta$), you will realize that the contribution to the force of any differential of the segment $\mathrm{d}\vec{F}$ points to the center of the arc. Due to the symmetry of the half circular arc, we can infer the contribution of the force along $\hat\imath$ will cancel out (consider the pairs of segment pieces at $\theta$ and $\pi-\theta$), but the contribution along $\hat\jmath$ will add up and therefore the resulting force on these segment should also be along $-\hat\jmath$.

In order to get the magnitude of the resulting on the arc you need to integrate $\mathrm{d}\vec{F}$ over $\theta$ from $0$ to $\pi$. By doing so, we can confirm our RHR deduction of the direction also. $$ \mathrm{d}\vec{F}_{\text{arc}}=I\,\left[R(\sin\theta\hat\imath-\cos\theta\hat\jmath)\mathrm{d}\theta\right]\times B\hat{k}=IRB\mathrm{d}\theta(\sin\theta\hat\imath-\cos\theta\hat\jmath)\times\hat{k}$$ $$=IRB\mathrm{d}\theta(-\sin\theta\hat{\jmath}-\cos\theta\hat\imath)=-IRB\mathrm{d}\theta(\cos\theta\hat\imath+\sin\theta\hat{\jmath})$$ $$\Rightarrow\vec{F}_\text{arc}=\int_{\theta=0}^{\theta=\pi}\mathrm{d}\vec{F}_{\text{arc}}=-IRB\int_{\theta=0}^{\theta=\pi}(\cos\theta\hat\imath+\sin\theta\hat{\jmath})\mathrm{d}\theta=-2IRB\hat\jmath$$

  • $\begingroup$ can u show me how can I do it using determinant way and cross product which I am confused onto how to do for the circular arc I got same value using determinant & cross product for F1 and F3 but can't seem to get it when I do it for the circular arc. Thanks alot. $\endgroup$ – Illustionisttt. Nov 24 '13 at 21:29
  • $\begingroup$ I included more steps in the end. Hope this clarifies it. Remember the cross products of the unit vectors: $\hat\imath\times\hat\jmath=\hat{k}$, $\hat\jmath\times\hat{k}=\hat\imath$, $\hat{k}\times\hat\imath=\hat\jmath$, and the cross-product is anti-commutative (changes sign when vectors are swapped). $\endgroup$ – Kevin Nov 24 '13 at 23:56

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