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# A bar magnet of magnetic moment $1.5\;J/T$ lies aligned with the direction of a uniform magnetic field of $0.22\;T$

$\begin{array}{1 1} 3.3\;J \\ 0.33\;J \\ 33\;J \\ 0.033\;J \end{array}$

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A)
Solution :
Given $M= 1.5 \;J/T$
$B= 0.22\;T$
$\theta _1 =0^{\circ}$
$\theta_2 =90^{\circ}$
Work done $w= -MB(\cos \theta_2 - \cos \theta_1)$
$\qquad= -1.5 \times 0.22 (\cos 90^{\circ} -\cos 0)$
$\cos 90^{\circ}=0$
and $\cos 0=1$
$\therefore W= -1.5 \times 0.22 \times -1$
$\qquad= 0.33 \;J$
Answer :$0.33\;J$