QJEEM23Phb0ef: JEE Main 2023 Physics - Class 12 JEEM23Phb0ef Solution: Two charges each of magnitude and separated by a distance of…

Given two charges each of magnitude $0.01 \,\text{C}$ and separated by a distance of $0.4 \,\text{mm} = 0.4 \times 10^{-3} \,\text{m}$ , we can calculate the dipole moment $\vec{p}$ :

$\vec{p} = q \cdot \vec{d} = (0.01 \,\text{C})(0.4 \times 10^{-3} \,\text{m}) = 4 \times 10^{-6} \,\text{Cm}$

The dipole is placed in a uniform electric field $\vec{E}$ of magnitude $10 \,\text{dyne/C}$ , which is equivalent to $10 \times 10^{-5} \,\text{N/C}$ .

The angle between the dipole moment and the electric field is $30^{\circ}$ .

The torque acting on the dipole can be calculated using the formula:

$\tau = pE \sin{\theta}$

Substituting the given values:

$\tau = (4 \times 10^{-6} \,\text{Cm})(10 \times 10^{-5} \,\text{N/C})\sin{30^{\circ}}$

$\tau = (4 \times 10^{-6} \,\text{Cm})(10^{-4} \,\text{N/C})(\frac{1}{2})$

$\tau = 2 \times 10^{-10} \,\text{Nm}$

The magnitude of the torque acting on the dipole is $2.0 \times 10^{-10} \,\text{Nm}$ .