A magnet of total magnetic moment $$10^{-2} \hat{i}$$ A m$$^2$$ is placed in a time varying magnetic field, $$B\hat{i}(\cos\omega t)$$ where $$B = 1$$ Tesla and $$\omega = 0.125$$ rad s$$^{-1}$$. The work done for reversing the direction of the magnetic moment at $$t = 1$$ second, is:

We have a magnetic dipole whose magnetic moment vector is given as $$\vec m = 10^{-2}\,\hat i\;{\rm A\,m^2}.$$

The external magnetic field is time-dependent and is specified by

$$\vec B(t)=B\cos\omega t\;\hat i,$$

where the numerical values are $$B = 1\ {\rm T}$$ and $$\omega = 0.125\ {\rm rad\,s^{-1}}.$$

At the instant $$t = 1\ {\rm s}$$ the field becomes

$$\vec B(1)=1\cos(0.125\times1)\;\hat i =\cos(0.125)\;\hat i\ {\rm T}.$$

The angle $$0.125$$ rad is small, so

$$\cos(0.125)\approx 0.9922.$$

Hence the field magnitude at this instant is approximately

$$B_1 = 0.9922\ {\rm T}.$$

The formula for the potential energy of a magnetic dipole in a magnetic field is

$$U = -\vec m\cdot\vec B.$$

Initially the dipole is parallel to the field, so the angle between $$\vec m$$ and $$\vec B$$ is $$0^\circ$$. Therefore

$$U_i = -mB_1\cos 0^\circ = -mB_1.$$

To reverse the dipole, we make it point opposite to the field; the angle becomes $$180^\circ$$. Then

$$U_f = -mB_1\cos 180^\circ = -mB_1(-1) = +mB_1.$$

The work done by an external agent equals the increase in potential energy:

$$W = U_f - U_i = \left(+mB_1\right) - \left(-mB_1\right) = 2mB_1.$$

Substituting $$m = 10^{-2}\ {\rm A\,m^2}$$ and $$B_1 = 0.9922\ {\rm T}$$, we get

$$W = 2 \times 10^{-2} \times 0.9922 = 0.019844\ {\rm J} \approx 0.02\ {\rm J}.$$

Hence, the correct answer is Option B.