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,
$\mu_{c} =i A=i \pi\left(\frac{l^{2}}{4 \pi^{2}}\right)$
$\mu_{c} =\frac{i l^{2}}{4 \pi}$...(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,
$\mu_{s} =i A=i \frac{1^{2}}{16}$
$\frac{\mu_{c}}{\mu_{s}} =\frac{\frac{i l^{2}}{4 \pi}}{\frac{il^{2}}{16}}=\frac{16}{4 \pi}=\frac{4}{\pi}$
Hence, the ratio of magnetic dipole moment of circular coil and square coil is $4: \pi$