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A square coil of sides $10 \;cm$ consists of 20 turns and carries a current of $12 A$ . The coil is suspended vertically and the normal to $30^{\circ}$ with the direction of a uniform horizontal magnetic field of magnitude $0.80\;T$ . What is the magnitude of torque experienced by the coil ?

$\begin{array}{1 1} 0.96\;Nm^{-1} \\ 0.96\;Nm^2 \\ 0.96\; N-m \\ 0.96\;Nm^{-2} \end{array} $

1 Answer

Solution :
Given , side of the square coil $=10\;cm=0.1 \;m$
Number of turns $n=20$
$I= 12 A$
$\theta= 30^{\circ}$
$B= 0.80T$
$T= NIAB \sin \theta$
$T= 20 \times 12 \times (10 \times 10^{-2})^2 \times 0.80 \times\sin 30^{\circ}$
$\qquad=20 \times 12 \times (10 \times 10^{-2})^2 \times 0.80 \times \large\frac{1}{2}$
$\qquad=0.96\; N-m $
Answer : $0.96\; N-m $
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