A) 2 : 1
B) 1 : 4
C) 4 : 1
D) 1 : 2
Correct Answer: B
Solution :
Magnetic induction at the centre of current carrying coil \[B=\frac{{{\mu }_{0}}\,ni}{2\,r}\] ...(i) Suppose the length of the wire be L. 1st case: For coil of one turn, let radius be \[{{r}_{1}}\] \[\therefore \] \[L=2\pi {{r}_{1}}\times n\] \[or{{r}_{1}}=\frac{L}{2\pi \times n}=\frac{L}{2\pi }\]\[(\because \,n=1)\] 2nd case : For coil of two turns, let radius be \[{{r}_{2}}\]. \[\therefore L=2\pi {{r}_{2}}\times n\] \[or{{r}_{2}}=\frac{L}{2\pi \times n}=\frac{1}{2\pi \times 2}\,\,\,(\because \,n=2)\] or \[{{r}_{2}}=\frac{{{r}_{1}}}{2}\] From Eq. (i), we have \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{n}_{1}}}{{{r}_{1}}}\times \frac{{{r}_{2}}}{{{n}_{2}}}\] or \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{1\times \frac{{{r}_{1}}}{2}}{{{r}_{1}}\times 2}\] \[\therefore \frac{{{B}_{1}}}{{{B}_{2}}}=\frac{1}{4}\]
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