1. Magnetic dipole moment of the coil

For a coil of $N$ turns, current $I$, and area $A$, the magnetic dipole moment is

$m = NIA$

Given:

• $N = 100$
• $I = 1\text{ mA} = 10^{-3}\text{ A}$
• $A = 5 \times 10^{-3}\text{ m}^2$

So,

$m = 100 \times 10^{-3} \times 5 \times 10^{-3}$

$m = 5 \times 10^{-4}\ \text{A m}^2$

2. Initial and final orientation

Potential energy of a magnetic dipole in a magnetic field is

$U = -mB\cos\theta$

where $\theta$ is the angle between magnetic moment and magnetic field.

• The plane of the coil is perpendicular to the magnetic field.
• Therefore the normal to the plane (and hence magnetic moment) is parallel to $\vec B$.
• So initially,

$\theta_i = 0^\circ$

After turning the coil through $90^\circ$, the magnetic moment also turns by $90^\circ$ relative to the field:

$\theta_f = 90^\circ$

3. Work done

Work done by an external agent in rotating the coil slowly is equal to increase in potential energy:

$W = U_f - U_i$

$W = (-mB\cos 90^\circ) - (-mB\cos 0^\circ)$

$W = (0) - (-mB) = mB$

Now,

$W = (5 \times 10^{-4})(0.20)$

$W = 1.0 \times 10^{-4}\ \text{J}$

4. Convert to microjoule

Since

$1\ \mu\text{J} = 10^{-6}\ \text{J}$

we get

$W = \frac{10^{-4}}{10^{-6}} = 100\ \mu\text{J}$

Final Answer

$\boxed{100}$