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  •  A long conducting wire having a current I flowing through it, is bent into a circular coil of t

 A long conducting wire having a current I flowing through it, is bent into a circular coil of \mathrm{N} turns. Then it is bent into a circular coil of \mathrm{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:

Option: 1

\mathrm{n: N}


Option: 2

n^{2}: \mathrm{N}^{2}


Option: 3

\mathrm{N^{2}: n^{2}}


Option: 4

\mathrm{N: n}


Answers (1)

best_answer

Length Remains Same.

\mathrm{\ell=N\left(2 \pi r_{1}\right)=n\left(2 \pi r_{2}\right)}
\mathrm{\frac{B_{1}}{B_{2}}=\frac{\left(N \frac{\mu_{0} I}{2 r_{1}}\right)}{\left(n \frac{\mu_{0} I}{2 r_{2}}\right)}=\frac{N}{n}\left(\frac{r_{2}}{r_{1}}\right)=\frac{N}{n}\left(\frac{N}{n}\right)}
\mathrm{\frac{B_{1}}{B_{2}}=\left(\frac{N}{n}\right)^{2}}

Posted by

Divya Prakash Singh

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