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

 

 

 

 

 

 

 

 

  • Option 1)

    \frac{m}{\pi }

  • Option 2)

    \frac{3m}{\pi }

  • Option 3)

    \frac{2m}{\pi }

  • Option 4)

    \frac{4m}{\pi }

Answers (1)

When square loop

magnetic diapole moment by square

M=Ia^{2}             (a= side of square )

When square convert into circle radius of circle is

2\pi r=4a

r=\frac{2a}{\pi }

magnetic moment by circle

M^{1} =I \times area

=I \times 4\pi r^{2}

=I\times \pi \times \frac{4a^{2}}{\pi r}

=\frac{I\times 4a^{2}}{\pi }

M^{1}=\frac{4m}{\pi }


Option 1)

\frac{m}{\pi }

Option 2)

\frac{3m}{\pi }

Option 3)

\frac{2m}{\pi }

Option 4)

\frac{4m}{\pi }

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