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A coil in the shape of an equilateral triangle of side 10cm lies in vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20 mT. The torque acting on the coil when current of 0.2 A passed through it and its plane becomes parallel to the magnetic field will be $\sqrt{x} \times 10^{- 5}$ Nm , the value of $x$ is ___

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answer is 3.

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Detailed Solution

$\begin{array}{l} area of equilateral triangle A = \frac{\sqrt{3}}{4} l^{2}, l = 10 cm = 10^{- 1} m \end{array}$

$B = 20 \times 10^{- 3} T$

$\begin{array}{l} Torque τ = BIAN \\ = Bi \frac{\sqrt{3} l^{2}}{4} i = 2 \times 10^{- 1} A \\ = 20 \times 10^{- 3} \times 2 \times 10^{- 1} \times \frac{\sqrt{3}}{4} \times {(10^{- 1})}^{2} \\ τ = \sqrt{3} \times 10^{- 5} Nm \end{array}$

$\Rightarrow x = 3$

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