A circular coil having 200 turns, $$2.5 \times 10^{-4} \text{ m}^2$$ area and carrying $$100\mu\text{A}$$ current is placed in a uniform magnetic field of 1T. Initially the magnetic dipole moment $$(\vec{M})$$ was directed along $$\vec{B}$$. Amount of work, required to rotate the coil through $$90°$$ from its initial orientation such that $$\vec{M}$$ becomes perpendicular to $$\vec{B}$$, is _______ $$\mu$$J.

A circular coil with 200 turns, area $$2.5 \times 10^{-4}$$ m² carrying $$100 \mu$$A current is in a uniform magnetic field of 1 T. We need the work to rotate it 90° from aligned to perpendicular orientation.

The magnetic dipole moment is $$M = NIA = 200 \times 100 \times 10^{-6} \times 2.5 \times 10^{-4}$$ which gives $$M = 200 \times 10^{-4} \times 2.5 \times 10^{-4} = 5 \times 10^{-6} \text{ A m}^2$$.

The potential energy of a magnetic dipole in a field is $$U = -MB\cos\theta$$. Initially ($$\theta = 0°$$), $$U_i = -MB\cos 0° = -MB$$ and finally ($$\theta = 90°$$), $$U_f = -MB\cos 90° = 0$$.

Therefore the work done is $$W = U_f - U_i = 0 - (-MB) = MB$$ which equals $$W = 5 \times 10^{-6} \times 1 = 5 \times 10^{-6} \text{ J} = 5 \text{ }\mu\text{J}$$.

The answer is $$\boxed{5}$$ $$\mu$$J.