# A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet such that B vector is in plane of the coil. if due to a current i in the triangle a torque torque acts on it, the side l of the triangle is

A coil is in the shape of an equilateral triangle so each side has length = l m, Magnetic field = B,
when a current (i) is passed through the coil having area A, then the magnetic dipole moment of the coil is given by
M=IA  where A is the area of the coil
$A=\frac{1}{2}×base×height=\frac{1}{2}×l×\mathrm{sin}60°=\frac{\sqrt{3}}{4}{l}^{2}$ [an equilateral triangle has an angle of 60 degree each]
therefore the magnetic dipole moment of the coil is
$M=iA=\frac{\sqrt{3}}{4}i$l2
The angle between the normal to the plane of the coil and the applied magnetic field is 90 degree.
Now the torque acting on the coil is given by

$\tau =MB\mathrm{sin}\theta =\frac{\sqrt{3}}{4}i{l}^{2}B\mathrm{sin}90=\frac{\sqrt{3}}{4}iB{l}^{2}$

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