Question

A circular loop of radius r, having N turns and carrying current I, is kept in the XY plane. It is then subjected to a uniform magnetic field B = Bx i + By j + Bz k. Obtain 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 , $\vec{M}= N\vec{A}\iota$

where A is cross sectional area of loop

so, A = πr² along z direction because loop placed in XY plane.

now, $\vec{M}=N(\pi r^2)\iota\hat{k}$

now, potential energy of the coil magnetic field system, $P.E=\vec{M}.\vec{B}$

so, P.E = $N(\pi r^2)\iota\hat{k}.(B_x\hat{i}+B_y\hat{j}+B_z\hat{k})$

P.E = $\pi r^2N\iota B_z$