Q140 / 160 · TS EAMCET 2015 · Physics

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
    $\frac{2}{\pi}$
  • C
    2
  • Correct answer
    D
    $\frac{4}{\pi}$

Solution

Dipole moments due to current carrying circular wire $M=i A$ Here, $$ \begin{aligned} \frac{M_1}{M_2} & =\frac{i \pi r^2}{i a^2}=\pi\left[\frac{\left(\frac{1}{2 \pi}\right)^2}{\left(\frac{1}{4}\right)^2}\right] \\ & =\pi \frac{16}{4 \pi^2}=\frac{4}{\pi} \end{aligned} $$

Hence, the correct answer is option: 'D'.

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