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  • When a long wire carrying a steady current is bent into a circular coil of one turn, the magnetic field at its centre is B. When the same wire carrying the same current is bent to form a circular c

When a long wire carrying a steady current is bent into a circular coil of one turn, the magnetic field at its centre is B. When the same wire carrying the same current is bent to form a circular coil of \mathrm{n}  turns of a smaller radius, the magnetic field at the centre will be

Option: 1

\mathrm{\frac{B}{n} }


Option: 2

\mathrm{n~ B }


Option: 3

\mathrm{\frac{B}{n^2} }


Option: 4

\mathrm{n^2 B}


Answers (1)

best_answer

\mathrm{B=\frac{\mu_0 I}{2 r}} . For a coil of  n turns, \mathrm{2 \pi r=n\left(2 \pi r^{\prime}\right)}
or   \mathrm{r^{\prime}=\frac{r}{n}}   where  \mathrm{r^{\prime}} is the radius of the coil of n turns.
\mathrm{ \therefore \quad B^{\prime}=\frac{n \mu_0 I}{2 r^{\prime}}=\frac{n \mu_0 I}{2 r / n}=n^2 B }

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Rishi

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