MHT CET · Physics · Magnetic Effects of Current
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.
- A \(4 \pi\)
- B \(\frac{4}{\pi}\)
- C \(\frac{2}{\pi}\)
- D \(2 \pi\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{\pi}\)
Step-by-step Solution
Detailed explanation
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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