Question

second loop of radius R having n turns and carrying current I is kept in the xy plane with it is then subjected to uniform magnetic field B equals to B XI + b y j + b jhatke obtain the expression for the magnetic potential energy of the coil magnetic field system​

Answer 1

A circular loop of radius r having N turns and carrying current i is kept in the XY plane.
so, magnetic moment ,
where A is cross sectional area of loop
so, A = πr² along z direction because loop placed in XY plane.
now,
now, potential energy of the coil magnetic field system,
so, P.E =
P.E =