A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m . If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:

Step-by-step solution

Let the side of square loop be a . The magnetic dipole is ∴ m = I a 2 a = m I Radius of circular loop r related to a as circumference of circular loop is equal to perimeter of square loop. 2 π r = 4 a = 4 m I r = 2 π m I ∴ New dipole moment m ' = π 2 π m I 2 I = 4 m π