Q139 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
A circular coil of N turns and diameter D carries current I. It is unwound and rewound to make another coil of diameter 2D, current I remaining the same. The ratio of magnetic moments of the original coil to the new coil is
- A$2:1$
- B$4:1$
- Correct answerC$1:2$
- D$1:4$
Solution
The length of the wire remains constant when unwound and rewound.
$L = N \pi D = N' \pi (2D)$
$N' = \dfrac{N}{2}$
The magnetic moment of a circular coil is given by $M = NIA$.
For the original coil:
$M_1 = NI \left( \dfrac{\pi D^2}{4} \right)$
For the new coil:
$M_2 = N' I \left( \dfrac{\pi (2D)^2}{4} \right) = \left( \dfrac{N}{2} \right) I (\pi D^2) = 2 NI \left( \dfrac{\pi D^2}{4} \right)$
Taking the ratio of the magnetic moments:
$\dfrac{M_1}{M_2} = \dfrac{1}{2}$
Answer: $1:2$
$L = N \pi D = N' \pi (2D)$
$N' = \dfrac{N}{2}$
The magnetic moment of a circular coil is given by $M = NIA$.
For the original coil:
$M_1 = NI \left( \dfrac{\pi D^2}{4} \right)$
For the new coil:
$M_2 = N' I \left( \dfrac{\pi (2D)^2}{4} \right) = \left( \dfrac{N}{2} \right) I (\pi D^2) = 2 NI \left( \dfrac{\pi D^2}{4} \right)$
Taking the ratio of the magnetic moments:
$\dfrac{M_1}{M_2} = \dfrac{1}{2}$
Answer: $1:2$
Hence, the correct answer is option: 'C'.