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 o

Question

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:

TestHub · Accepted Answer

As in both cases, length of the wire will remain the same, therefore

$l = (2 π R) N = (2 π r) n \Rightarrow \frac{R}{r} = \frac{n}{N}$

Now magnetic field due to single circular coil is given by, $B = (\frac{μ_{0} I}{2 r})$ .

Therefore, required ratio of total magnetic field wil be,

$\frac{B_{1}}{B_{2}} = \frac{N (\frac{μ_{0} I}{2 R})}{n (\frac{μ_{0} I}{2 r})} = (\frac{N^{2}}{n^{2}})$

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