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

Q: 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

c $m = I{l^2}$ $2\pi r = 4l$ ${{{\text{m}}^\prime } = \frac{{I4{\ell ^2}}}{\pi }}$ ${{\text{r}} = \frac{{2\ell }}{\pi }}$ ${{{\text{m}}^\prime } = \frac{{I4{\ell ^2}}}{\pi }}$ ${\pi {r^2} = \frac{{\pi 4{\ell ^2}}}{{{\pi ^2}}} = \frac{{4{\ell ^2}}}{\pi }}$ ${{{\text{m}}^\prime } = \frac{4}{\pi }\,{\text{m}}}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$\frac {m}{\pi }$
B$\frac {3m}{\pi }$
C$\frac {4m}{\pi }$
D$\frac {2m}{\pi }$

JEE MAIN 2019, Medium

STD 11 Science
NEET
STD 12 - 4. Moving charges and magnetism
Physics

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