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Question:

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 :

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