QJEEM24Ph6e7e: JEE Main 2024 Physics - Class 12 JEEM24Ph6e7e Solution: A coil having 100 turns, area of , carrying current of is…

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 $100 \mu \mathrm{J}$ .