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:

Question

Accepted Answer

4m/π. Square side a gives m = I a^2 with perimeter 4a. Same wire forms a circle: 2*pi*r = 4a => r = 2a/pi, area = pi r^2 = 4a^2/pi. So m_circle = I*(4a^2/pi) = 4m/pi.

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:

Solution

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