InterviewSolution
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InterviewSolution
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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. |
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Answer» 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))` |
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