A short bar magnet of magnetic moment 0.4,JT^ | vedclass

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Question

(B) The potential energy $U$ of a magnetic dipole in a magnetic field is given by the formula $U = -\overrightarrow{M} \cdot \overrightarrow{B} = -MB

Vedclass · Accepted Answer

$-0.064\,J$

Vedclass · Answer

(B) The potential energy $U$ of a magnetic dipole in a magnetic field is given by the formula $U = -\overrightarrow{M} \cdot \overrightarrow{B} = -MB \cos 	heta$.For stable equilibrium,the magnetic moment $\overrightarrow{M}$ must be aligned with the magnetic field $\overrightarrow{B}$,which means the angle $	heta = 0^{\circ}$.Substituting the given values $M = 0.4\,JT^{-1}$ and $B = 0.16\,T$ into the formula:$U = -(0.4\,JT^{-1})(0.16\,T) \cos(0^{\circ})$$U = -0.064 	imes 1 = -0.064\,J$.Therefore,the potential energy in stable equilibrium is $-0.064\,J$.

$A$ short bar magnet of magnetic moment $0.4\,JT^{-1}$ is placed in a uniform magnetic field of $0.16\,T$. The magnet is in stable equilibrium when the potential energy is

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