A long conducting wire having a current I flowing through it, is bent into a circular coil of N turns. Then it is bent into a circular coil of n turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is

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A long conducting wire having a current I flowing through it, is bent into a circular coil of N turns. Then it is bent into a circular coil of n turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is

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n²: N²

N : n

N²: n²

n : N

answer is C.

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Detailed Solution

At centre of circular coil $B = \frac{μ_{0} Ni}{2 R}$
1st case $ℓ = N 2 {πR}_{1} \Rightarrow R_{1} = \frac{ℓ}{2 πN}$
$\begin{array}{l} B_{1} = \frac{μ_{0} NI}{2 ℓ} (2 πN) \\ B_{1} = \frac{μ_{0} {πN}^{2}}{ℓ} I \end{array}$
2nd case $B_{2} = \frac{μ_{0} nI}{2 \frac{ℓ}{2 πn}} = \frac{μ_{0} {πn}^{2} I}{ℓ}$
$\frac{B_{1}}{B_{2}} = \frac{N^{2}}{n^{2}}$