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Magnetic dipole in a uniform field
Appears in
Concepts tested here
- Potential energy of a dipole
- Work and torque on a magnetic dipole
All Questions
2012 AIPMT-MAINS 1 question
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A magnetic needle suspended parallel to a magnetic field requires $\sqrt3$ J of work to turn it through $60^\circ$. The torque needed to maintain the needle in this position will beWork done $= U_f - U_i = -MB\cos60^\circ - (-MB\cos0^\circ) = MB\left(1 - \frac{1}{2}\right) = \frac{MB}{2}$
$\sqrt3 = \frac{MB}{2} \Rightarrow MB = 2\sqrt3$ J
Torque needed: $\tau = MB\sin60^\circ = 2\sqrt3\times\frac{\sqrt3}{2}$
$\tau = 3$ J
2011 AIPMT-MAINS 1 question
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A short bar magnet of magnetic moment 0.4 J $T^{-1}$ is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium when the potential energy isPotential energy of a magnetic dipole in a field: $U = -\vec{M}\cdot\vec{B} = -MB\cos\theta$
Stable equilibrium corresponds to minimum potential energy, which occurs when the magnetic moment is along the field: $\theta = 0^\circ$.
$U = -MB\cos0^\circ = -0.4\times0.16\times1$
$U = -0.064$ J