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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

A short bar magnet of magnetic moment m = 0.32 J T–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

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प्रश्न

A short bar magnet of magnetic moment m = 0.32 J T–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

  1. stable, and
  2. unstable equilibrium?

What is the potential energy of the magnet in each case?

संख्यात्मक
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उत्तर

Given: Moment of the bar magnet, M = 0.32 J T−1

External magnetic field, B = 0.15 T

(a) The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle θ, between the bar magnet and the magnetic field is 0°.

The potential energy of the system:

U = `vecM * vecB`

 = −MB cos θ

= −0.32 × 0.15 × cos 0°

= −0.32 × 0.15 × 1

= − 0.048 J

(b) The bar magnet is oriented 180° to the magnetic field. Hence, it is in unstable equilibrium.

θ = 180°

Potential energy of the system:

U = `vecM * vecB`

= −MB cos θ

= −MB cos 180°

= −0.32 × 0.15 × (−1)

= +0.048 J

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अध्याय 5: Magnetism and Matter - EXERCISES [पृष्ठ १५२]

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एनसीईआरटी Physics Part I and II [English] Class 12
अध्याय 5 Magnetism and Matter
EXERCISES | Q 5.2 | पृष्ठ १५२
एनसीईआरटी Physics Part I and II [English] Class 12
अध्याय 5 Magnetism and Matter
Exercise | Q 5.4 | पृष्ठ २००

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