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Q. The magnitude of torque experienced by a square coil of side $12 \,cm$ which consists of $25$ turns and carries a current $10\, A $ suspended vertically and the normal to the plane of coil makes an angle of $30^{\circ}$ with the direction of a uniform horizontal magnetic field of magnitude $0.9 \,T$ is

Moving Charges and Magnetism

Solution:

$\tau =NIA\, B\, sin \, \theta$
Here, $N=25$,
$I=10\, A$,
$B=0.9 \, T$,
$\theta =30^{\circ}$
$A=a^{2}=12 \times 10^{-2} \times 12\times 10^{-2}$
$144 \times 10^{-4}\, m^{2}$
$\therefore \tau=25 \times 10 \times = 144\times 10^{-4}\times 0.9\times sin \, 30^{\circ}$
$=16.\, N\,m$