A magnetic dipole experiences a torque of \( 80\sqrt{3} \) N m when placed in a uniform magnetic field in such a way that the dipole moment makes an angle of \( 60^\circ \) with the magnetic field. The potential energy of the dipole is:
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Remember the formulas for the torque \( \tau = MB \sin \theta \) and potential energy \( U = -MB \cos \theta \) of a magnetic dipole in a uniform magnetic field. Use the given torque and angle to find the product \( MB \), and then use this value to calculate the potential energy at the same angle.
The torque \( \tau \) experienced by a magnetic dipole in a uniform magnetic field \( \vec{B} \) is expressed as:\[\tau = \vec{M} \times \vec{B} = MB \sin \theta\]Here, \( M \) denotes the magnitude of the magnetic dipole moment, \( B \) represents the magnitude of the magnetic field, and \( \theta \) is the angle between \( \vec{M} \) and \( \vec{B} \).Given \( \tau = 80\sqrt{3} \) N m and \( \theta = 60^\circ \):\[80\sqrt{3} = MB \sin 60^\circ\]\[80\sqrt{3} = MB \left( \frac{\sqrt{3}}{2} \right)\]Solving for \( MB \):\[MB = \frac{80\sqrt{3} \times 2}{\sqrt{3}} = 160\]The potential energy \( U \) of the magnetic dipole within the uniform magnetic field is defined by:\[U = -\vec{M} \cdot \vec{B} = -MB \cos \theta\]Substituting \( MB = 160 \) and \( \theta = 60^\circ \):\[U = -(160) \cos 60^\circ\]\[U = -160 \left( \frac{1}{2} \right)\]\[U = -80 \text{ J}\]
