A square coil of side $10\; cm$ 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 $30^o$ with the direction of a uniform horizontal magnetic field of magnitude $0.80 \;T$. What is the magnitude of torque (in $N\;m$) experienced by the coil?
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Length of a side of the square coil, $l=10 \,cm =0.1\, m$
Current flowing in the coil, $I=12\, A$
Number of turns on the coil, $n=20$
Angle made by the plane of the coil with magnetic field, $\theta=30^{\circ}$
Strength of magnetic field, $B=0.80\, T$
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
$\tau=n B L A \sin \theta$
$\Rightarrow l \times l=0.1 \times 0.1=0.01\, m^{2}$
$\therefore \tau=20 \times 0.8 \times 12 \times 0.01 \times \sin 30^{\circ}$
$=0.96 \,Nm$
Hence, the magnitude of the torque experienced by the coil is $0.96 \,N m$
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