1. Magnetic potential energy of a current loop

For a coil in a uniform magnetic field, $U = -\vec{M}\cdot\vec{B} = -MB\cos\theta$ where:

• $M$ = magnetic dipole moment of the coil
• $B$ = magnetic field
• $\theta$ = angle between $\vec{M}$ and $\vec{B}$

The work required to rotate the coil slowly from initial angle $\theta_i$ to final angle $\theta_f$ is the increase in potential energy: $W = U_f - U_i$

2. Calculate magnetic dipole moment

For a coil of $N$ turns, $M = NIA$

Given:

• $N = 200$
• $I = 100\,\mu\text{A} = 100 \times 10^{-6}\,\text{A} = 10^{-4}\,\text{A}$
• $A = 2.5 \times 10^{-4}\,\text{m}^2$

So, $M = 200 \times 10^{-4} \times 2.5 \times 10^{-4}$ $M = 200 \times 2.5 \times 10^{-8}$ $M = 500 \times 10^{-8}$ $M = 5 \times 10^{-6}\,\text{A m}^2$

3. Initial and final energies

Initially, $\vec{M}$ is along $\vec{B}$, so $\theta_i = 0^\circ$ $U_i = -MB\cos 0^\circ = -MB$

Finally, $\vec{M}$ is perpendicular to $\vec{B}$, so $\theta_f = 90^\circ$ $U_f = -MB\cos 90^\circ = 0$

Thus, $W = U_f - U_i = 0 - (-MB) = MB$

4. Substitute values

Given $B = 1\,\text{T}$, $W = M B = 5 \times 10^{-6} \times 1 = 5 \times 10^{-6}\,\text{J}$

Convert to microjoule: $5 \times 10^{-6}\,\text{J} = 5\,\mu\text{J}$

5. Final answer

$\boxed{5}$

So the required work is $5\,\mu\text{J}$.