A circular loop and a square loop are formed from two wires of same length and cross section. Same current is passed through them. Then, the ratio of their dipole moments is

Question

A circular loop and a square loop are formed from two wires of same length and cross section. Same current is passed through them. Then, the ratio of their dipole moments is (A) 4 (B) (2/π) (C) 2 (D) (4/π)

Tardigrade Admin · Accepted Answer

Dipole moments due to current carrying circular wire M=i A Here, (M1/M2) =(i π r2/i a2)=π[(((1/2 π))2/((1)4)2)] =π (16/4 π2)=(4/π)

Q. A circular loop and a square loop are formed from two wires of same length and cross section. Same current is passed through them. Then, the ratio of their dipole moments is

4

$\frac{2}{π}$

2

$\frac{4}{π}$

Dipole moments due to current carrying circular
wire $M = i A$
Here, $\frac{M _{1}}{M _{2}} = \frac{i π r ^{2}}{i a ^{2}} = π [\frac{( \frac{1}{2 π} ) ^{2}}{( \frac{1}{4} ) ^{2}}]$
$= π \frac{16}{4 π ^{2}} = \frac{4}{π}$

Q. A circular loop and a square loop are formed from two wires of same length and cross section. Same current is passed through them. Then, the ratio of their dipole moments is

4

π2​

2

π4​

Dipole moments due to current carrying circular wire M=iA Here, M2​M1​​=ia2iπr2​=π[(41​)2(2π1​)2​] =π4π216​=π4​

$\frac{2}{π}$

$\frac{4}{π}$

Dipole moments due to current carrying circular
wire $M = i A$
Here, $\frac{M _{1}}{M _{2}} = \frac{i π r ^{2}}{i a ^{2}} = π [\frac{( \frac{1}{2 π} ) ^{2}}{( \frac{1}{4} ) ^{2}}]$
$= π \frac{16}{4 π ^{2}} = \frac{4}{π}$