Magnetics 6 Question 9
9. A magnet of total magnetic moment
$10^{-2} \hat{\mathbf{i}} A-m^{2}$ is placed in a time varying magnetic field, $B \hat{\mathbf{i}}$ $(\cos \omega t$ ), where $B=1 T$ and $\omega=0.125 rad / s$. The work done for reversing the direction of the magnetic moment at $t=1 s$ is
(2019 Main, 10 Jan I)
(a) $0.01 J$
(b) $0.007 J$
(c) $0.014 J$
(d) $0.028 J$
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Solution:
- Work done in reversing dipole is
$$ W=2 M B $$
where, $M$ =magnetic dipole moment
and
$$ \begin{aligned} & =10^{-2} A-m^{2} \\ B & =\text { external field } \\ & =B \cos \omega t=1 \times \cos (0.125 \times 1) \\ & =\cos \left(7^{\circ}\right)=0.992 \end{aligned} $$
Substituting these values, we get,
$$ \begin{aligned} W & =2 \times 10^{-2} \times 0.992 \\ & =0.0198 J \end{aligned} $$
which is nearest to $0.014 J$
