To find the work done in turning the coil through $$90^{\circ}$$, we first need to understand the concept of torque on a current-carrying loop in a magnetic field and how work done relates to the change in potential energy of the system. The potential energy (U) of a magnetic dipole in a magnetic field is given by:

$$U = - \vec{M} \cdot \vec{B}$$

where:

• $\vec{M}$ is the magnetic moment of the coil, and
• $\vec{B}$ is the magnetic field.

For a coil with $N$ turns, carrying current $I$, and with an area $A$, the magnetic moment $\vec{M}$ is defined as:

$$M = NI \cdot A$$

Given that the coil has $$100$$ turns, carries a current of $$1 \, \mathrm{mA} = 1 \times 10^{-3} \, \mathrm{A}$$, and the area of the coil is $$5 \times 10^{-3} \, \mathrm{m}^2$$, we can calculate its magnetic moment as follows:

$$M = 100 \cdot 1 \times 10^{-3} \cdot 5 \times 10^{-3} = 0.5 \times 10^{-3} \, \mathrm{Am}^2$$

Since the coil is initially placed such that its plane is perpendicular to the magnetic field (i.e., the angle $ \theta = 0^{\circ} $), and then it is turned through $90^{\circ}$, the initial and final angles ($\theta_i$ and $\theta_f$) of the coil with respect to the magnetic field are $0^{\circ}$ and $90^{\circ}$ respectively. This means the initial potential energy ($U_i$) and final potential energy ($U_f$) of the system are:

$$U_i = - M B \cos(\theta_i)$$

$$U_f = - M B \cos(\theta_f)$$

Given that $B = 0.20 \, \mathrm{T}$, $\theta_i = 0^{\circ} \, (\cos(0) = 1)$, and $\theta_f = 90^{\circ} \, (\cos(90^{\circ}) = 0)$, the potential energies are:

$$U_i = - 0.5 \times 10^{-3} \cdot 0.20 \cdot 1 = - 1 \times 10^{-4} \, \mathrm{J}$$

$$U_f = - 0.5 \times 10^{-3} \cdot 0.20 \cdot 0 = 0 \, \mathrm{J}$$

The work done ($W$) is equal to the change in potential energy:

$$W = U_f - U_i$$

$$W = 0 - (- 1 \times 10^{-4}) = 1 \times 10^{-4} \, \mathrm{J}$$

Therefore, the work done in turning the coil through $90^{\circ}$ is $$100 \mu \mathrm{J}$$.