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A circular loop and a square loop are formed from the same wire and the same current is passed through them. Find the ratio of their dipole moments.
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2920 Upvotes
Verified Answer
The correct answer is:
$\frac{4}{\pi}$
Suppose, the length of wire is $l$. When, wire formed in circular loop then radius of loop,
$$
r=\frac{l}{2 \pi}
$$
$[\because l=2 \pi r]$
Magnet dipole, $M_{1}=i A$
$$
\begin{array}{l}
=i \pi r^{2}=i \pi \times\left(\frac{l}{2 \pi}\right)^{2} \\
=i \times \frac{l^{2}}{4 \pi}
\end{array}
$$
When, wire formed in square loop then the side of loop,
$$
a=\frac{l}{4}
$$
Magnet dipole, $M_{2}=I A$
$$
\begin{aligned}
&=i \times a^{2} \\
&=i \times \frac{l^{2}}{16} \\
\frac{M_{1}}{M_{2}} &=\frac{4}{\pi}
\end{aligned}
$$
$$
r=\frac{l}{2 \pi}
$$
$[\because l=2 \pi r]$
Magnet dipole, $M_{1}=i A$
$$
\begin{array}{l}
=i \pi r^{2}=i \pi \times\left(\frac{l}{2 \pi}\right)^{2} \\
=i \times \frac{l^{2}}{4 \pi}
\end{array}
$$
When, wire formed in square loop then the side of loop,
$$
a=\frac{l}{4}
$$
Magnet dipole, $M_{2}=I A$
$$
\begin{aligned}
&=i \times a^{2} \\
&=i \times \frac{l^{2}}{16} \\
\frac{M_{1}}{M_{2}} &=\frac{4}{\pi}
\end{aligned}
$$
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