I'm trying to redo the textbook example in magnetostatics. The current carrying loop is affected by a magnetic flux in the ax→ direction. While trying to find the torque on sides ab and cd for figure c and d, I noticed that they used sine and cosine. Where did the sine and cosine come from?
Best Answer
in your textbook this formula
$\tau=\vec r \times \vec F$
where $\vec r$ is sepration vector joining origin to the force vector, is used to find torque due to forces $\vec F_{ab}$ and $\vec F_{cd}$ respectivaly
torque due to $\vec F_{ab}$
$\vec \tau _{ab}=\vec r_{ab}\times \vec F_{ab}=\left|\dfrac{W}{2}\right|\left( \vec a_{x} sin\theta - \vec a_{y} cos\theta \right)\times |ILB|(\vec{-a_{y}})=-\left|\dfrac{ILBW}{2}\right| sin\theta \ \ \vec a_{z} $
torque due to $\vec F_{cd}$
$\vec \tau _{cd}=\vec r_{cd}\times \vec F_{cd}=\left|\dfrac{W}{2}\right|\left( -\vec a_{x} sin\theta + \vec a_{y} cos\theta \right)\times |ILB|(\vec{a_{y}})=-\left|\dfrac{ILBW}{2}\right| sin\theta \ \ \vec a_{z} $
so, those sines and cosines are due to components of position vector after rotation of loop
alternative method:
background :
the torque is given by
$\vec{\tau}=\vec M \times \vec B$ where
$\vec M=n I\vec A$ is magnetic moment vector and $B$ is magnetic field vector
but ,
in your question: after the loop has rotated by angle $\theta$ about z-axis
now area vector or normal vector of the loop (and hence, the magnetic moment vector) is pointing in oblique direction (in x y plane) as shown in figure (d) of your textbook
$\vec M= |IWL|\left(cos\theta \vec a_{x}+sin\theta \vec a_{y}\right)\ \ \ $ (inside || i've magntiude of that vector)
and $\vec B = |B|\ \vec a_{x}$
therefore,
$\implies \tau=-|IBWL|sin\theta \vec a_{z}\tag{1}$
thus, total torque acting on rectangular loop is given by above expression out of which half part is contributed by side $ab$ and half by side $cd$