A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III and IV, arrange them in the decreasing order of potential energy

Question

A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III and IV, arrange them in the decreasing order of potential energy (A) I > III > II > IV (B) I > II > III > IV (C) I > IV > II > III (D) III > IV > I > II

Tardigrade Admin · Accepted Answer

U = - MB = - Mcos θ Here, M = magnetic moment of the loop 9 = angle between M and B U is maximum when θ = 180° and minimum when θ = 0°. So, as θ decreases from 180° to 0° its PE also decreases.

Q. A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III and IV, arrange them in the decreasing order of potential energy

I > III > II > IV

I > II > III > IV

I > IV > II > III

III > IV > I > II

U = - MB = - $M cos θ$
Here, M = magnetic moment of the loop 9 = angle between M and B
U is maximum when $θ$ = 180 $^{\circ}$ and minimum when $θ$ = 0 $^{\circ}$ .
So, as $θ$ decreases from 180 $^{\circ}$ to 0 $^{\circ}$ its PE also decreases.

Q. A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III and IV, arrange them in the decreasing order of potential energy

I > III > II > IV

I > II > III > IV

I > IV > II > III

III > IV > I > II

U = - MB = -Mcosθ Here, M = magnetic moment of the loop 9 = angle between M and B U is maximum when θ = 180∘ and minimum when θ = 0∘. So, as θ decreases from 180∘ to 0∘ its PE also decreases.

U = - MB = - $M cos θ$
Here, M = magnetic moment of the loop 9 = angle between M and B
U is maximum when $θ$ = 180 $^{\circ}$ and minimum when $θ$ = 0 $^{\circ}$ .
So, as $θ$ decreases from 180 $^{\circ}$ to 0 $^{\circ}$ its PE also decreases.