JEE Main 2020 · 9 January · morning · Single correct · +4 / −1
An electric dipole of moment
p⃗ = ( - î - 3ĵ + 2k̂ ) × 10⁻²⁹ C.m is
at the origin (0, 0, 0). The electric field due to this dipole at
r⃗ = + î + 3ĵ + 5k̂ (note that r⃗ .p⃗ = 0 ) is parallel to :
- (a) ( + î + 3ĵ - 2k̂ )
- (b) ( + î - 3ĵ - 2k̂ )
- (c) ( - î + 3ĵ - 2k̂ )
- (d) ( - î - 3ĵ + 2k̂ )
Answer and solution
Answer (a) ( + î + 3ĵ - 2k̂ )
Solution
Physics sir · 4 steps
r⃗ · p⃗ = 0 means the equatorial line, and there the field points along -p⃗.
- General dipole field. For a dipole at the origin, the field at position r⃗ is given below. It has a part along r and a part along -p⃗.
E⃗ = 1/4πε₀ (3(p⃗ · r) r - p⃗)/r³
- Check the dot product. r⃗ · p⃗ = (1)(-1) + (3)(-3) + (5)(2) = -1 - 9 + 10 = 0. So the first term vanishes and E⃗ = -1/4πε₀p⃗/r³.
- Reverse the dipole. The field is along -p⃗. Flipping every sign of -î - 3ĵ + 2k̂ gives î + 3ĵ - 2k̂.
- So. The field is parallel to (+î + 3ĵ - 2k̂), which is option (a).
The idea to keep. On the equatorial line of a dipole, the field always points opposite to the dipole moment. Whenever r⃗ · p⃗ = 0, reverse p⃗ and you have the field direction.
Answer from the NTA final answer key. Solution by PrepWiser.