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A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $\vec{B}$ is in the plane of the coil. If due to a current $i$ in the triangle a torque $\tau$ acts on it, the side $l$ of the triangle is -
A
$\dfrac{2}{\sqrt{3}}\left(\dfrac{\tau}{Bi}\right)$
B
$\dfrac{1}{\sqrt{3}}\dfrac{\tau}{Bi}$
C
$2\left(\dfrac{\tau}{\sqrt{3}\,Bi}\right)^{1/2}$
D
$\dfrac{2}{\sqrt{3}}\left(\dfrac{\tau}{Bi}\right)^{1/2}$
Detailed Solution
Torque on a current loop: $\tau = MB\sin\theta$, where $M = iA$ is the magnetic moment and $\theta$ is the angle between $\vec{M}$ and $\vec{B}$.
$\vec{B}$ is in the plane of the coil, so $\vec{M}$ (normal to the coil) is perpendicular to $\vec{B}$: $\theta = 90^\circ$ and $\tau = iAB$.
Area of an equilateral triangle of side $l$: $A = \dfrac{\sqrt{3}}{4}l^2$
$\tau = i\left(\dfrac{\sqrt{3}}{4}l^2\right)B$
$l^2 = \dfrac{4\tau}{\sqrt{3}\,Bi}$
$l = 2\left(\dfrac{\tau}{\sqrt{3}\,Bi}\right)^{1/2}$
$\vec{B}$ is in the plane of the coil, so $\vec{M}$ (normal to the coil) is perpendicular to $\vec{B}$: $\theta = 90^\circ$ and $\tau = iAB$.
Area of an equilateral triangle of side $l$: $A = \dfrac{\sqrt{3}}{4}l^2$
$\tau = i\left(\dfrac{\sqrt{3}}{4}l^2\right)B$
$l^2 = \dfrac{4\tau}{\sqrt{3}\,Bi}$
$l = 2\left(\dfrac{\tau}{\sqrt{3}\,Bi}\right)^{1/2}$