Question:
Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field \(B_1\), at its centre when a current \(I\) passes through it. The ratio \(B_1 : B_2\) is:
Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field \(B_1\), at its centre when a current \(I\) passes through it. The ratio \(B_1 : B_2\) is:
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For length L bent into N turns, \(r = L/(2\pi N)\) and \(B = N\mu_0I/(2r) = \mu_0\pi N^2I/L\), so B scales as \(N^2\) at fixed L and I.
Updated On: Nov 11, 2026
- \(1:1\)
- \(1:3\)
- \(1:9\)
- \(9:1\)
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The Correct Option is C
Solution and Explanation
The printed question omits the description of the second wire; in the original NTA paper the other wire (same length L) is bent into a coil of three identical loops adjacent to each other carrying the same current I, and \(B_2\) is the field at the centre of that coil. This is the only reading that matches the options.
Single loop: \(2\pi r_1 = L \Rightarrow r_1 = \dfrac{L}{2\pi}\), \(B_1 = \dfrac{\mu_0 I}{2r_1}\).
Three loops: \(3\times2\pi r_2 = L \Rightarrow r_2 = \dfrac{r_1}{3}\), \(B_2 = \dfrac{3\mu_0I}{2r_2} = \dfrac{9\mu_0I}{2r_1} = 9B_1\).
\(B_1 : B_2 = 1 : 9\).
Answer: (3) \(1:9\)
Single loop: \(2\pi r_1 = L \Rightarrow r_1 = \dfrac{L}{2\pi}\), \(B_1 = \dfrac{\mu_0 I}{2r_1}\).
Three loops: \(3\times2\pi r_2 = L \Rightarrow r_2 = \dfrac{r_1}{3}\), \(B_2 = \dfrac{3\mu_0I}{2r_2} = \dfrac{9\mu_0I}{2r_1} = 9B_1\).
\(B_1 : B_2 = 1 : 9\).
Answer: (3) \(1:9\)
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