JEE MAIN - Physics (2023 - 13th April Morning Shift - No. 9)
Two charges each of magnitude $$0.01 ~\mathrm{C}$$ and separated by a distance of $$0.4 \mathrm{~mm}$$ constitute an electric dipole. If the dipole is placed in an uniform electric field '$$\vec{E}$$' of 10 dyne/C making $$30^{\circ}$$ angle with $$\vec{E}$$, the magnitude of torque acting on dipole is:
$$4 \cdot 0 \times 10^{-10} ~\mathrm{Nm}$$
$$1.5 \times 10^{-9} ~\mathrm{Nm}$$
$$1.0 \times 10^{-8} ~\mathrm{Nm}$$
$$2.0 \times 10^{-10} ~\mathrm{Nm}$$
Explanation
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 $$2.0 \times 10^{-10} \,\text{Nm}$$.
$$\vec{p} = q \cdot \vec{d} = (0.01 \,\text{C})(0.4 \times 10^{-3} \,\text{m}) = 4 \times 10^{-6} \,\text{Cm}$$
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 $$2.0 \times 10^{-10} \,\text{Nm}$$.
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