5.4 A short bar magnet of magnetic moment m = 0.32 JT^–1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

<p><strong>5.4 </strong>Moment of the bar magnet, M = 0.32 J/T</p><p>Magnetic field, B = 0.15 T</p><p>The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle<span contenteditable="false"> <math> <mi>θ</mi> </math> </span> , between the bar magnet and the magnetic field is 0<span contenteditable="false"> <math> <mo>°</mo> </math> </span>.</p><p>Potential energy of the system = -MBcos<span contenteditable="false"> <math> <mi>θ</mi> </math> </span> = - 0.32<span contenteditable="false"> <math> <mo>×</mo> <mn>0.15</mn> <mo>×</mo> <mrow> <mrow> <mi mathvariant="normal">cos</mi> </mrow> <mo>?</mo> <mrow> <mn>0</mn> <mo>°</mo> </mrow> </mrow> </math> </span> = -4.8 <span contenteditable="false"> <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> </math> </span>J</p><p>When the bar magnet is oriented 180<span contenteditable="false"> <math> <mo>°</mo> </math> </span> to the magnetic field, it becomes unstable equilibrium.</p><p>Potential energy = - MBcos<span contenteditable="false"> <math> <mi>θ</mi> </math> </span> = - 0.32 <span contenteditable="false"> <math> <mo>×</mo> <mn>0.15</mn> <mo>×</mo> <mrow> <mrow> <mi mathvariant="normal">cos</mi> </mrow> <mo>?</mo> <mrow> <mn>180</mn> <mo>°</mo> </mrow> </mrow> </math> </span>= 4.8<span contenteditable="false"> <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> </math> </span> J</p>

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