A small current-carrying loop is located at a distance `r` from a long straight conductor with current `I`. The magnetic moment of the loop is equal to `p_(m)`. Find the magnitude and direction of the force vector applied to the loop if the vector `p_(m)`
(a) is parallel to the stratight conductor,
(b) is oriented along the radius vector `r`,
(c) coincides in direction with the magnetic field produced by the current `I` the point where the loop is located.

Explanation

Due to the straight conductor, `B_(varphi) = (mu_(0) I)/(2pi r)`
We use the formula, `vec(F) = (vec(p_(m)). vec(grad)) vec(B)`
(a) The vector `vec(p_(m))` is parallel to the staright conductor.
`vec(F) = p_(m) (del)/(del Z) vec(B) = 0`,
because netiher the direction nor the magnitude of `vec(B)` depends on `z`
(b) The vector `vec(p_(m))` is oriented along the radius vector `vec(r)`
`vec(F) = p_(m) (del)/(del r) vec(B)`
The direction of `vec(B)` at `r + dr` is parallel to the direction at `r`. Thus only the `varphi` component of `vec(F)` will survive.
`F_(varphi) = p_(m) (del)/(del r) (mu_(0) I)/(2pi r) = - (mu_(0) I p_(m))/(2pi^(2))`
(c) The vector `vec(p_(m))` coincides in direction with the magnetic field, produced by the conductor carrying current `I`
`vec(F) = p_(m) (del)/(r del varphi) (mu_(0) I)/(2pi) vec(e_(varphi)) = (mu_(0) I p_(m))/(2pi r^(2)) (del vec(e_(varphi)))/(del varphi)`
So, `vec(F) = (mu_(0) I p_(m))/(2pi r^(2)) vec(e_(r))` As, `(del vec(e_(varphi)))/(del_(varphi)) = -vec(e_(r))`