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

<p>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:</p> <p>$U = - \vec{M} \cdot \vec{B}$</p> <p>where:</p> <ul> <li>$\vec{M}$ is the magnetic moment of the coil, and</li> <li>$\vec{B}$ is the magnetic field.</li> </ul> <p>For a coil with $N$ turns, carrying current $I$, and with an area $A$, the magnetic moment $\vec{M}$ is defined as:</p> <p>$M = NI \cdot A$</p> <p>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:</p> <p>$$M = 100 \cdot 1 \times 10^{-3} \cdot 5 \times 10^{-3} = 0.5 \times 10^{-3} \, \mathrm{Am}^2$$</p> <p>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:</p> <p>$U_i = - M B \cos(\theta_i)$</p> <p>$U_f = - M B \cos(\theta_f)$</p> <p>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:</p> <p>$U_i = - 0.5 \times 10^{-3} \cdot 0.20 \cdot 1 = - 1 \times 10^{-4} \, \mathrm{J}$</p> <p>$U_f = - 0.5 \times 10^{-3} \cdot 0.20 \cdot 0 = 0 \, \mathrm{J}$</p> <p>The work done ($W$) is equal to the change in potential energy:</p> <p>$W = U_f - U_i$</p> <p>$W = 0 - (- 1 \times 10^{-4}) = 1 \times 10^{-4} \, \mathrm{J}$</p> <p>Therefore, the work done in turning the coil through $90^{\circ}$ is $100 \mu \mathrm{J}$.</p>