I was reading an answer about torque acting on a rectangular current carrying loop kept in a uniform magnitude field B. Force acting on each sides is $F_1$, $F_2$, $F_3$, $F_4$. It's written here : Force $F_2$, $F_4$ acting on these sides has same magnitude $F' = Blb \sin(90-\theta) = Blb \cos\theta$. And $F1$ and $F3$, have $F=IlB\sin90° = IlB$.

(where $l$=length; $b$=breadth) I understand $F_1$ and $F_3$. But I don't get how the magnitude of forces of $F_2$, $F_4$ become $Blb \cos\theta$ . Which law says that? Am i missing some knowledge?

If you need the diagram or the book text for more details , please ask for it.

  • $\begingroup$ $\uparrow$ Which answer? $\endgroup$ – Qmechanic Mar 29 '15 at 17:45

The force that a current carrying wire experiences in a magnetic field is known as the Lorentz force, in differential form: $$ d\mathbf{F} = \int_{wire} I d\mathbf{l}\times\textbf{B}$$.

$d\mathbf{l}$ is in the direction of the current.

Since $I$ is uniform along the wire, the integration is just replaced by $\cdot L$, where $L$ is the length of the wire.

The magnitude of the cross product $| \mathbf{A}\times \mathbf{B} | = |\mathbf{A}|\cdot |\mathbf{B}|\cdot \sin(\phi)$ where $\phi$ is the angle between $\mathbf{A}$ and $\mathbf{B}$.

I am assuming $\theta$ is the angle by which the rectangular loop is deflected with respect to, say, the xy plane.

If sides 2 and and 4 are the ones perpendicular to the xy plane, and $\mathbf{B}$ is in the $z$ direction, then the angle between $d\mathbf{l}$ and $\mathbf{B}$ is $90 - \theta$.

enter image description here

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  • $\begingroup$ But how is F dependent on B and length l and breadth b both in the first case while for F1 and F3. It is dependent on current I , length, and B. I think instead of Blb it should be BIb . $\endgroup$ – pblead26 Aug 6 '14 at 5:40
  • $\begingroup$ you have written the same thing twice in the last sentence, what do you mean? $\endgroup$ – SuperCiocia Aug 6 '14 at 16:19
  • $\begingroup$ I see the confusion. In my text book, it is written " B l b " (BEE, el, bee), but I think it should be " B I b " ( BEE, AAI, bee ). I think the printer must have printed l(el) instead of I(aai). Dont you think? $\endgroup$ – pblead26 Aug 6 '14 at 16:24
  • $\begingroup$ ok let me think about that, and consider using LaTeX since it would make it much clearer: $Blb$ and $BIb$ $\endgroup$ – SuperCiocia Aug 6 '14 at 16:27
  • $\begingroup$ okay, sorry. I have to learn how to use that. $\endgroup$ – pblead26 Aug 6 '14 at 16:28

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