A rectangular coil of length $0 .12\, m$ and width $0.1\, m$ having $50$ turns of wire is suspended vertically in a uniform magnetic field of strength $0.2\,\, Weber/m^2.$ The coil carries a current of $2 \,\,A.$ If the plane of the coil is inclined at an angle of $30^o $ with the direction of the field, the torque required to keep the coil in stable equilibrium will be .......$Nm$
AIPMT 2015, Medium
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The required torque is $\tau=N I A B \sin \theta$
where $N$ is the number of turns in the coil, $I$ is the current through the coil, $B$ is the uniform magnetic field, $A$ is the area of the coil and $\theta$ is the angle between the direction of the magnetic field and nomal to the plane of the coil. Here, $N=50, I=2\, \mathrm{A},$
$A=0.12\, \mathrm{m} \times 0.1\, \mathrm{m}=0.012\, \mathrm{m}^{2}$
$B=0.2\, \mathrm{Wb} / \mathrm{m}^{2} \text { and } \theta=90^{\circ}-30^{\circ}=60^{\circ}$
$\therefore \quad \tau =(50)(2\, \mathrm{A})\left(0.012\, \mathrm{m}^{2}\right)\left(0.2\, \mathrm{Wb} / \mathrm{m}^{2}\right) \sin 60^{\circ} $
$=0.20\, \mathrm{N} \mathrm{m}$
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