A wire of length L carrying current i is bent into circular loop with (i) one turn (ii) n turns. Find the ratio of magnetic induction at centre in above two cases.

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TestHub · Accepted Answer

The magnetic field at the center of a circular loop is $B = \frac{μ _{0} ni}{2 R}$ .

For one turn, $L = 2 π R_{1}$ , so $R_{1} = \frac{L}{2 π}$ . $B_{1} = \frac{μ _{0} i}{2 R _{1}} = \frac{μ _{0} i}{2 ( L /2 π )} = \frac{μ _{0} iπ}{L}$ .

For n turns, $L = n (2 π R_{2})$ , so $R_{2} = \frac{L}{2 nπ}$ . $B_{2} = \frac{μ _{0} ni}{2 R _{2}} = \frac{μ _{0} ni}{2 ( L /2 nπ )} = \frac{μ _{0} n ^{2} iπ}{L}$ .

The ratio is $\frac{B _{1}}{B _{2}} = \frac{μ _{0} iπ / L}{μ _{0} n ^{2} iπ / L} = \frac{1}{n ^{2}}$ . So, $1 : n^{2}$ .

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