Question

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. \(4:1\)
• C. \(1:2\)
• D. \(1:4\)

Hint:

Wire length is fixed, so doubling the diameter halves the number of turns.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The magnetic moment of a coil is \(M=NIA\), where \(A\) is the area of each turn. The total wire length stays the same when the coil is rewound.

Step 2: Find the new number of turns:
Original wire length \(=N\times\pi D\). New coil has diameter \(2D\), so \(N'\times\pi(2D)=N\pi D\), giving \(N'=\dfrac N2\).

Step 3: Compare magnetic moments:
\(M_1=NI\pi\dfrac{D^2}{4}\). \(M_2=\dfrac N2I\pi\dfrac{(2D)^2}{4}=\dfrac N2I\pi D^2\).
\[ \dfrac{M_1}{M_2}=\dfrac{N\pi D^2/4}{N\pi D^2/2}=\dfrac12 \]
So the ratio is \(1:2\). Option C.

Step 4: Why the other options are wrong.
2:1 and 4:1 would come from ignoring the change in turns. 1:4 would arise by multiplying area by 4 and forgetting turns also change.

Final Answer:
The ratio of magnetic moments is 1 : 2. \[ \boxed{\text{(C) }1:2} \]