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A square current carrying coil of edge length L.the magnetic field on the coil given by Bvector=B0y/L icap +B0x/L jcap where B0 is a positive constant.(A is area of coil).

1) If the coil is free to rotate about x axis torque on coil is given by 1/2iAB0icap

2)if the coil is free to rotate about y axis torque on the coil is given by -1/2iAB0j cap

3) resultant force on coil is zero

4)equation for torque μ vector cross B vector where μ is magnetic moment of coil is not valid on coil if any of the side is fixed as axis

$Torqueonsquareloopinmagneticfield=BIA\mathrm{cos}\theta where\theta istheanglebetween\phantom{\rule{0ex}{0ex}}magneticfieldandareavector\phantom{\rule{0ex}{0ex}}TheforcewillbeBIL\mathrm{sin}\theta andisequalandopposite.Sotheresu\mathrm{tan}atforcewillbezero.\phantom{\rule{0ex}{0ex}}REGARDS\phantom{\rule{0ex}{0ex}}$

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