The current density in a cylindrical wire of radius r = 4 . 0   mm is 1 . 0 × 10 6   A 2   m 2 . The current through the outer portion of the wire between radial distances r 2 and r is x π   A ; where x is

Step-by-step solution

Current = J × Area = J × π R 2 - R 2 2 = J π 3 R 2 4 = 10 6 × π × 3 × 4 × 10 - 3 4 2 ⇒ π × 3 × 16 4 = 12 π   A Hence, the value of x = 12 .