Magnetics Ques 139
- 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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Answer:
Correct Answer: 139.(c)
Solution:
- Work done in reversing dipole is
$ W=2 M B $
where, $M$ =magnetic dipole moment
$ =10^{-2} A-m^{2} $
and $B =\text { external field } $
$ =B \cos \omega t=1 \times \cos (0.125 \times 1) $
$=\cos \left(7^{\circ}\right)=0.992$
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$
BETA
