Question 194: Magnetism and Matter

Question

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:

Accepted Answer

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

Answer

$${m \over \pi }$$

Answer

$${{3m} \over \pi }$$

Answer

$${{2m} \over \pi }$$

Answer

$${{4m} \over \pi }$$

Magnetism and Matter

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:

Choose the correct answer: