A magnet of total magnetic moment 10 ^ - 2 ^ i A m ^ 2 is placed in...

As we know that, work done in rotating a magnetic dipole in magnetic field is,

$W = MB ( \cos \left({\theta}_{1}\right) - \cos \left({\theta}_{2}\right) ) . . . ( 1 )$

Now in the above given question we have, Magnetic Moment ( M ) = {10}^{- 2}\overset{^}{i}A{m}^{2},Magnetic Field= B\overset{^}{i}\cos \left(\omegat\right) with $B = 1T$ , $\omega = 0 . 125rad{s}^{- 1}$ and $t = 1s$

Let initial angle $( {\theta}_{1} ) = {0}^{∘}$ so on reversing the direction of magnetic moment final angle $( {\theta}_{2} ) = {180}^{∘}$

Now, substituting all the values in equation $( 1 )$ we get,

$⇒ W = {10}^{- 2} \times ( 1 ) \cos \left(0 . 125 \times 1\right)\left(\cos \left({0}^{∘}\right) - \cos \left({180}^{∘}\right)\right)$

$⇒ W = {10}^{- 2}\cos \left(0 . 125\right)\left(1 - \left(- 1\right)\right) ( ∵ \cos \left({0}^{∘}\right) = 1\text{and}\cos \left({180}^{∘}\right) = - 1 )$

$⇒ W = 2 \times {10}^{- 2} \times 0 . 99$

$⇒ W \approx 0 . 02J$

Therefore, the work done in reversing the direction of the magnetic moment is $0 . 02J$.