If a bar magnet in magnetic moment m is deflected from an angle θ in a uniform magnetic field of induction B, the energy stored in reversing the direction is : Now lets assume that the bar magnetic is initially placed parallel to the field the initial potential energy will be Ui=−M.B=−MBcos0=−MB Now when it is deflected by an angle θ then Uf=−M.B=−MBcosθ ΔU=Ui−Uf ΔU=−MB(1−cosθ)
example
Potential energy of a current loop in a magnetic field
A planar coil of area 7m2 carrying an anti-clockwise current 2A is placed in an extemal magnetic field B=(0.2i^+0.2j^−0.3k^), such that the normal to the plane is along the line(3i^−5j^+4k^) Area A=7m2, Normal n^=501(3i^−5j^+4k^) ≃71(3i^−5j^+4k^) ∴ area A=(3i^−5j^+4k^)m2 Magnetic field B=(0.2i^+0.2j^−0.3k^)T Current I=2A ∴ magnetic moment M=IA=(6i^−10j^+8k^)Am2 ∣M∣≃14Am2(2×7=14) Potential energy U=−M.B =−[1.2−2−2.4]=3.2J