An iron rod of volume 10 - 3 m 3 and relative permeability 1000 is placed as core in a solenoid with 10 turns cm - 1 . If a current of 0 . 5 A is passed through the solenoid, then the magnetic moment of the rod will be :

Step-by-step solution

Given, Volume of iron rod is, V = 10 - 3   m 3 Relative permeability, μ r = 1000 No of turns per unit length of solenoid is, n = N l = 10   turns   cm - 1 = 1000   turns   m - 1 Current through the solenoid is, i = 0 . 5   A Magnetic moment of an iron core solenoid is given by, M = ( μ r - 1 )   N i A ⇒ M = μ r - 1 N i V l = μ r - 1 N l i V ⇒ M = 1000 - 1 × 1000 × 0 . 5 × 10 - 3 ⇒ M = 499 . 5 ≈ 500   A   m 2