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See diagram.

let us say the question is about an electric dipole or about a magnetic dipole. It is very similar for magnetic or electric dipole.

Let an electric dipole of charges +q at point A and -q at point B be attached by a thin weightless dielectric rod AB of length 2a. The electric dipole moment is 2 a q = p. The thin rod id held fixed at a pivot O in the center of AB. It can rotate freely in the plane of electric field E and the dipole axis.

Let there be an electric field E uniform in that region. Let the electric dipole line joining them make an angle with the electric field E. Then there is a force F on +q in the direction of E, and F = + qE. There is a force on the charge -q (coulomb's force) F2 = - F = -q E.

Since these two forces are not collinear, they do not cancel. There is a torque around the fixed point O. Because F and -F try to rotate the rod AB in the same direction.

magnitude of Torque = sum of arm length * force

= a Sin θ * q E + a Sin θ * q E

= 2 a q E Sin θ = p E Sin θ

In vector notation, we have

= 2 a q E Sin θ

The dipole and field are in x-y plane and torque is perpendicular to them is parallel to the z axis, passing through O.

let us say the question is about an electric dipole or about a magnetic dipole. It is very similar for magnetic or electric dipole.

Let an electric dipole of charges +q at point A and -q at point B be attached by a thin weightless dielectric rod AB of length 2a. The electric dipole moment is 2 a q = p. The thin rod id held fixed at a pivot O in the center of AB. It can rotate freely in the plane of electric field E and the dipole axis.

Let there be an electric field E uniform in that region. Let the electric dipole line joining them make an angle with the electric field E. Then there is a force F on +q in the direction of E, and F = + qE. There is a force on the charge -q (coulomb's force) F2 = - F = -q E.

Since these two forces are not collinear, they do not cancel. There is a torque around the fixed point O. Because F and -F try to rotate the rod AB in the same direction.

magnitude of Torque = sum of arm length * force

= a Sin θ * q E + a Sin θ * q E

= 2 a q E Sin θ = p E Sin θ

In vector notation, we have

**= Sigma***T***= a q E Sin θ***r X F***+ (- a) ( - q E)***k**k*= 2 a q E Sin θ

**=***k**p X E*The dipole and field are in x-y plane and torque is perpendicular to them is parallel to the z axis, passing through O.