A coil having 100 turns, area of $$5 \times 10^{-3} \text{ m}^2$$, carrying current of $$1 \text{ mA}$$ is placed in uniform magnetic field of $$0.20 \text{ T}$$ such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through $$90°$$ is ___________ $$\mu\text{J}$$.

We need to find the work done in turning a current-carrying coil through 90° in a uniform magnetic field. First, recall that the magnetic moment of a coil is $$M = NIA$$, where $$N$$ is the number of turns, $$I$$ is the current, and $$A$$ is the area. The potential energy of a magnetic dipole in a field is $$U = -MB\cos\theta$$, and the work done is the change in potential energy, given by $$W = U_f - U_i$$.

Using the specified values, the magnetic moment is $$M = NIA = 100 \times 1 \times 10^{-3} \times 5 \times 10^{-3}$$, which simplifies to $$M = 100 \times 5 \times 10^{-6} = 5 \times 10^{-4}\,\text{A·m}^2$$.

The plane of the coil is initially perpendicular to the magnetic field, so the magnetic moment, which is normal to the coil’s plane, is parallel to the field. Therefore, $$\theta_i = 0°$$. After rotating through 90°, the angle becomes $$\theta_f = 90°$$.

Applying the expression for work, we have $$W = U_f - U_i = (-MB\cos 90°) - (-MB\cos 0°) = 0 - (-MB) = MB$$.

Substituting $$M = 5 \times 10^{-4}\,\text{A·m}^2$$ and $$B = 0.20\,\text{T}$$ yields $$W = 5 \times 10^{-4} \times 0.20 = 1 \times 10^{-4}\,\text{J} = 100\,\mu\text{J}$$.

The correct answer is 100 $$\mu$$J.