A very long cylindrical wire is carrying a current I o distributed uniformly over its cross-section area. O is the centre of the cross-section of the wire and the direction of current in into the plane of the figure. The value of ∫ A B B → . dI → along the path AB (from A to B ) is

Step-by-step solution

According to Ampere's circuital law ∮ B → . dl → = μ 0 I net For the closed path OAB ∮ B → . dl → = ∫ O A B → . dl → + ∫ A B B → . dl → +   ∫ B O B → . dl → = μ 0 - I 0 6 Since the magnetic field is perpendicular to the paths OA and BO ∫ A B B → . dl → = - μ 0 I 0 6