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A square coil of side $10 \mathrm{~cm}$ consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of $30^{\circ}$ with the direction of a uniform horizontal magnetic field of magnitude $0.80 \mathrm{~T}$. What is the magnitude of torque experienced by the coil?
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Verified Answer
Given, $A=10 \times 10=100 \mathrm{~cm}^2=10^{-2} \mathrm{~m}^2$ $\mathrm{N}=20, \mathrm{I}=12 \mathrm{~A}, \mathrm{~B}=0.8 \mathrm{~T}, \theta=30^{\circ} \Rightarrow \sin \theta=0.5$
As we know, $\tau=\mathrm{NIAB} \sin \theta$
$$
\begin{aligned}
&=20 \times 12 \times 10^{-2} \times 0.8 \times 0.5 \\
&=96 \times 10^{-2} \mathrm{Nm}
\end{aligned}
$$
As we know, $\tau=\mathrm{NIAB} \sin \theta$
$$
\begin{aligned}
&=20 \times 12 \times 10^{-2} \times 0.8 \times 0.5 \\
&=96 \times 10^{-2} \mathrm{Nm}
\end{aligned}
$$
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