Manipal Entrance Test 2015 Physics Paper

The potential energy of circular loop in the magnetic field = - M B cos θ, where θ is the angle between normal to plane of coil and magnetic field B.
Initial potential energy,
Ui=−NiAB‌cos‌0°=−NiAB
When coil is turned through 180°, therefore final potential energy,
Uf=−NiAB‌cos‌180°=NiAB
Required work, W = gain in potential energy
=Uf−Ui =NiAB−(−NiAB) =2NiAB
Here, N =100, i = 0.1A, B = 1.5 Wb/m ‌2 and radius r = 0.05 m
∴A=πr2=3.14×(0.05)2 =7.85×10−3m2
∴Work, W=2×100×0.1×7.85×10−3×1.5
⇒W=0.2355J