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 :
mπ
4mπ
2mπ
3mπ
Solution:
The correct option is B 4mπ m=NIA=1×I×a² here a = side of square Now, 4a=2πr r=2a/π For circular loop m'=1×I×πr²=1×I×π×(2a/π)² m'=4mπ