A) nB
B) n2B
C) 2nB
D) 2n2B
Correct Answer: B
Solution :
Magnetic field at the center of single turn loop \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\pi i}{r}\], magnetic field at the center of n-turn loop \[{{B}_{n}}=\left( \frac{{{\mu }_{0}}}{4\pi }.\frac{2\pi i}{r/n} \right)\times n\Rightarrow {{B}_{n}}={{n}^{2}}B\]
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