Question

A circular coil and a square coil is prepared from two identical metal wires and a current is passed through it. Ratio of magnetic dipole moment associated with circular coil to that of square coil is

• A. $\frac{2}{\pi }$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $\frac{4}{\pi }$

Answer 1

Answer: (D)

Let $l$ be the length of metal wire.
When wire is bent into a circular coil of radius $r$, then
$r=\frac{l}{2\pi }$
$\therefore$ Area, $A=\pi {\left(\frac{l}{2\pi }\right)}^2=\pi \frac{l^2}{4{\pi }^2}$
$\therefore$ Magnetic dipole moment associated with circular coil,
${\mu }_c=iA=i\pi \left(\frac{l^2}{4{\pi }^2}\right)$
${\mu }_c=\frac{i{l}^2}{4\pi }$...(i)
When metal wire is bent into a square coil then side of square
$a=\frac{l}{4}$
$\therefore$ Area, $A=a^2=\frac{l^2}{16}$
$\therefore$ Magnetic dipole moment associated with square coil,
${\mu }_s=iA=i\frac{1^2}{16}$
$\frac{{\mu }_c}{{\mu }_s}=\frac{\frac{i{l}^2}{4\pi }}{\frac{i{l}^2}{16}}=\frac{16}{4\pi }=\frac{4}{\pi }$
Hence, the ratio of magnetic dipole moment of circular coil and square coil is $4:\pi$