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 ______.

Options

• ${\displaystyle \frac{2}{\pi }}$

• ${\displaystyle \frac{\pi }{2}}$

• π

• ${\displaystyle \frac{4}{\pi }}$

Answer 1

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 ${\displaystyle \underline{\frac{4}{\pi }}}$.

Explanation:

Let I be the length of metal wire.

When wire is bent into a circular coil of radius r, then

r = ${\displaystyle \frac{l}{2\pi }}$

∴ Area, A = ${\displaystyle \pi {\left(\frac{l}{2\pi }\right)}^2=\pi \frac{l^2}{4{\pi }^2}}$

∴ Magnetic dipole moment associated with circular coil,

${\displaystyle {\mu }_{\text{c}}-\text{iA}=\text{i}\pi \left(\frac{l^2}{4{\pi }^2}\right)}$

${\displaystyle {\mu }_{\text{c}}=\frac{{\text{il}}^2}{4\pi }}$     ....(i)

When metal wire Is bent into a square coil then side of square

a = ${\displaystyle \frac{l}{4}}$

∴ Area, A = ${\displaystyle {\text{a}}^2=\frac{l^2}{16}}$

∴ Magnetic dipole moment associated with square coil,

${\displaystyle {\mu }_{\text{s}}=\text{i A}=\text{i} \frac{l^2}{16}}$

∴ ${\displaystyle \frac{{\mu }_{\text{c}}}{{\mu }_{\text{s}}}=\frac{\frac{\text{i}{l}^2}{4\pi }}{\frac{\text{i}{l}^2}{16}}}$

${\displaystyle =\frac{16}{4\pi }=\frac{4}{\pi }}$

Hence, the ratio of magnetic dipole moment of circular coil and square coil Is 4 : π.