Q139 / 150 · MHT CET 2025 (23 Apr Shift 2) · Physics

The relation between magnetic moment $(\mathrm{M})$ of a current carrying circular coil and length (L) of the wire used is

  • A
    $M \alpha L$
  • B
    $\mathrm{M} \alpha \frac{1}{\mathrm{~L}}$
  • Correct answer
    C
    $M \alpha L^2$
  • D
    $\mathrm{M} \propto \frac{1}{\mathrm{~L}^2}$

Solution

The magnetic moment of a circular coil with N turns carrying current I is expressed as $M = NIA$, where $A = \pi r^2$ is the area per turn.

Given a fixed wire length $L = N(2\pi r)$, the radius is determined by $r = \frac{L}{2\pi N}$.

Substituting into the area formula yields $A = \pi \left(\frac{L}{2\pi N}\right)^2 = \frac{L^2}{4\pi N^2}$.

Thus, the magnetic moment becomes $M = NI \cdot \frac{L^2}{4\pi N^2} = \frac{I L^2}{4\pi N}$.

For constant I and N, $M \propto L^2$, indicating direct proportionality to the square of the wire length.

The correct choice is $\boxed{\text{C}}$.

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

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