QJEEM19Phb8eb: JEE Main 2019 Physics - Class 12 JEEM19Phb8eb Solution: A square loop is carrying a steady current I and the magnitude…

Let the given square loop has side $a$ , then its magnetic dipole moment will be

$m=I a^2$

When square is converted into a circular loop of radius $r$ ,


Then, wire length will be same in both area,

$\Rightarrow 4 a=2 \pi r \Rightarrow r=\frac{4 a}{2 \pi}=\frac{2 a}{\pi}$

Hence, area of circular loop formed is, $A^{\prime}=\pi r^2$

$=\pi\left(\frac{2 a}{\pi}\right)^2=\frac{4 a^2}{\pi}$

Magnitude of magnetic dipole moment of circular loop will be

$m^{\prime}=I A^{\prime}=I \frac{4 a^2}{\pi}$

Ratio of magnetic dipole moments of both shapes is,

$\begin{aligned}\frac{m^{\prime}}{m}=\frac{I \cdot \frac{4 a^2}{\pi}}{I a^2}=\frac{4}{\pi} \\\\\Rightarrow m^{\prime} =\frac{4 m}{\pi}(\mathrm{A}-\mathrm{m})\end{aligned}$