JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 13)
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:
$${m \over \pi }$$
$${{3m} \over \pi }$$
$${{2m} \over \pi }$$
$${{4m} \over \pi }$$
Explanation
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 $r$,
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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,
$$ \begin{aligned} \frac{m^{\prime}}{m}=\frac{I \cdot \frac{4 a^2}{\pi}}{I a^2}=\frac{4}{\pi} \\\\ \Rightarrow m^{\prime} =\frac{4 m}{\pi}(\mathrm{A}-\mathrm{m}) \end{aligned} $$
$$ m=I a^2 $$
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,
$$ \begin{aligned} \frac{m^{\prime}}{m}=\frac{I \cdot \frac{4 a^2}{\pi}}{I a^2}=\frac{4}{\pi} \\\\ \Rightarrow m^{\prime} =\frac{4 m}{\pi}(\mathrm{A}-\mathrm{m}) \end{aligned} $$
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