A) \[nB\]
B) \[{{n}^{2}}B\]
C) \[2nB\]
D) \[2{{n}^{2}}B\]
Correct Answer: B
Solution :
The magnetic field at the centre of circular coil is \[B=\frac{{{\mu }_{0}}i}{2r}\] where\[r=\]radius of circle\[=\frac{l}{2\pi }\] \[(\because l=2\pi r)\] \[\therefore \] \[B=\frac{{{\mu }_{0}}i}{2}\times \frac{2\pi }{l}\] \[=\frac{{{\mu }_{0}}i\pi }{l}\] ...(i) When wire of length I bents into a circular loops of n turns, then \[l=n\times 2\pi r\] \[\Rightarrow \] \[r=\frac{l}{n\times 2\pi }\] Thus, new magnetic field \[B=\frac{{{\mu }_{0}}ni}{2r}=\frac{{{\mu }_{2}}ni}{2}\times \frac{n\times 2\pi }{l}\] \[=\frac{{{\mu }_{0}}i\pi }{l}\times {{n}^{2}}\] \[={{n}^{2}}B\] [from Eq. (i)]
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