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 \[=\frac{l}{2\pi }\] \[\therefore \] \[B=\frac{{{\mu }_{0}}i}{2}\times \frac{2\pi }{l}\] \[(\because l=2\pi r)\] \[\Rightarrow \] \[B=\frac{{{\mu }_{0}}i\pi }{l}\] ?.(i) When wire of length \[l\] 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 \[B'=\frac{{{\mu }_{0}}ni}{2r'}=\frac{{{\mu }_{0}}ni}{2}\times \frac{n\times 2\pi }{l}\] \[=\frac{{{\mu }_{0}}i\pi }{l}\times {{n}^{2}}={{n}^{2}}B\] [from Eq.(i)]
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