JEE Main 2023 · 13 April · morning · Single correct · +4 / −1
Two charges each of magnitude 0.01 C and separated by a distance of 0.4 mm constitute an electric dipole. If the dipole is placed in an uniform electric field 'E⃗' of 10 dyne/C making 30° angle with E⃗, the magnitude of torque acting on dipole is:
- (a) 4 · 0 × 10⁻¹⁰ Nm
- (b) 1.5 × 10⁻⁹ Nm
- (c) 1.0 × 10⁻⁸ Nm
- (d) 2.0 × 10⁻¹⁰ Nm
Answer and solution
Answer (d) 2.0 × 10⁻¹⁰ Nm
Solution
Physics sir · 3 steps
Turn dyne/C into N/C first, then torque is just pE sin θ.
- Dipole moment. p = q × 2a = 0.01 × 0.4 × 10⁻³ = 4 × 10⁻⁶ C m.
- Field in SI. 1 dyne = 10⁻⁵ N, so E = 10 × 10⁻⁵ = 10⁻⁴ N C⁻¹.
τ = pE sin θ = 4 × 10⁻⁶ × 10⁻⁴ × sin 30°
- So. τ = 4 × 10⁻¹⁰ × 1/2 = 2.0 × 10⁻¹⁰ N m, which is option (d).
The idea to keep. Torque on a dipole is pE sin θ, largest at 90° and zero when the dipole is along the field. Always bring CGS units like dyne into SI before multiplying.
Answer from the NTA final answer key. Solution by PrepWiser.