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NEET PHYSICSEasy

A short bar magnet of magnetic moment 0.4 J T−10.4 \text{ J T}^{-1} is placed in a uniform magnetic field of 0.16 T0.16 \text{ T}. The magnet is in stable equilibrium when the potential energy is:

A

0.064 J

B

zero

C

−0.082 J

D

−0.064 J

Step-by-Step Solution

  1. Formula: The potential energy (UU) of a magnetic dipole with magnetic moment m\mathbf{m} in a uniform magnetic field B\mathbf{B} is given by the dot product: U=−m⋅B=−mBcos⁡θU = -\mathbf{m} \cdot \mathbf{B} = -mB \cos \theta where θ\theta is the angle between the magnetic moment vector and the magnetic field vector .
  2. Stable Equilibrium: The magnet is in stable equilibrium when the potential energy is at its minimum. This occurs when the magnetic moment is aligned parallel to the magnetic field, i.e., θ=0∘\theta = 0^\circ .
  3. Calculation:
  • Magnetic Moment (mm) = 0.4 J T−10.4 \text{ J T}^{-1}
  • Magnetic Field (BB) = 0.16 T0.16 \text{ T}
  • Angle (θ\theta) = 0∘0^\circ U=−(0.4)×(0.16)×cos⁡0∘U = -(0.4) \times (0.16) \times \cos 0^\circ U=−0.064×1=−0.064 JU = -0.064 \times 1 = -0.064 \text{ J}
  1. Conclusion: The potential energy in the stable equilibrium position is −0.064 J-0.064 \text{ J}.
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