Question

If a single-turn circular coil ( of radius \(R_{1}\)) carrying a current \(I\) is rewound into a coil of \(N\) turns ( of radius \(R_{2}\)) using the same total length of wire and carrying the same current \(I\), the ratio of the magnetic field at the centre of the new coil \((B_{2})\) to that of the original coil \((B_{1})\) is

• A. \(N\)
• B. \(N^{2}\)
• C. \(\frac{1}{N^{2}}\)
• D. \(2N\)

Hint:

The wire length is fixed, so N times 2 pi R2 equals 2 pi R1. Then compare the fields at the centre.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The same length of wire is used, so \(2\pi R_{1}=N\cdot2\pi R_{2}\), giving \(R_{2}=\dfrac{R_{1}}{N}\).

Step 2: Field at the centre
For one turn \(B_{1}=\dfrac{\mu_{0}I}{2R_{1}}\). For N turns \(B_{2}=\dfrac{N\mu_{0}I}{2R_{2}}\).

Step 3: Calculation
\(B_{2}=\dfrac{N\mu_{0}I}{2R_{1}/N}=N^{2}\cdot\dfrac{\mu_{0}I}{2R_{1}}\)

Step 4: Ratio
\(\dfrac{B_{2}}{B_{1}}=N^{2}\)

Final Answer:
The ratio of the new field to the old field is N squared. \[ \boxed{\text{(B) }N^{2}} \]