The figure below shows the north and south poles of a permanent magnet in which $n$ turn coil of area of cross-section $A$ is resting, such that for a current $ i$ passed through the coil, the plane of the coil makes an angle $\theta $ with respect to the direction of magnetic field $B.$ If the plane of the magnetic field and the coil are horizontal and vertical respectively, the torque on the coil will be
A$\tau = niAB\cos \theta $
B$\tau = niAB\sin \theta $
C$\tau = niAB$
D
None of the above, since the magnetic field is radial
Medium
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A$\tau = niAB\cos \theta $
a (a)Plane of coil is having angle $\theta $ with the magnetic field.
$\therefore \;\tau = MB\sin (90 - \theta )$ or $\tau = niAB\cos \theta $ [As $M =niA$]
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