A small bar magnet placed with its axis
at 30o with an external field of 0.06 T
experiences a torque of 0.018 Nm. The
minimum work required to rotate it from its
stable to unstable equilibrium position is :
Solution
Torque on a bar magnet :
<br><br>$\tau = MB\,\sin \theta$
<br><br>Here, $\theta$ = 30º, I = 0.018 N-m, B = 0.06 T
<br><br>$0.018 = M \times 0.06 \times 0.5$
<br><br>$\Rightarrow M = 0.6\,A{m^2}$<br><br>$W = {U_f} - {U_i}$<br><br>$= MB(\cos {\theta _i} - \cos {\theta _f})$<br><br>$= 0.6 \times 0.06(1 - ( - 1))$<br><br>$= 7.2 \times {10^{ - 2}}\,J$
About this question
Subject: Physics · Chapter: Magnetism · Topic: Bar Magnet and Magnetic Field
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