Correct answer is 100

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:

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

where:

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

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

$U_i=-MB\cos ({\theta }_i)$

$U_f=-MB\cos ({\theta }_f)$

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 $100\mu \mathrm{J}$.

Correct answer option is 5

$\begin{array}{rl}& W=U_f-U_i=(-MB\cos 90)-(-MB\cos 0)\\ & \Rightarrow W=MB=NiAB=5\mu \mathrm{J}\end{array}$

Correct answer option is 5

$\begin{array}{rl}& iLB=mg\text{ and }L=\frac{AR}{\rho }\\ & \begin{array}{rl}\therefore B& =\frac{mg\rho }{iAR}\\ & =\frac{0.5\times 10\times 2\times {10}^{-6}}{2\times 10\times {10}^{-6}\times 1}\\ & =0.5\mathrm{T}\end{array}\end{array}$

Correct answer is 50

$\begin{array}{rl}& {\overrightarrow{F}}_{BC}+{\overrightarrow{F}}_{DA}=0\\ & {\overrightarrow{F}}_{AB}=ilB=0.5\times 0.5(5)=1.25\mathrm{N}\times 0.2=0.25\mathrm{N}\\ & {\overrightarrow{F}}_{CD}=0.5\times 0.5(6)=1.5\times 0.2=0.3\mathrm{N}\\ & F_{\text{net }}=0.05\mathrm{N}\\ & =50\mathrm{mN}\end{array}$