A) 0.2
B) 0.4
C) 0.6
D) 0.8
Correct Answer: D
Solution :
Given : \[{{B}_{1}}=0.2\] tesla As number of turns are doubled \[{{N}_{1}}=2\]and \[{{N}_{2}}=2\times 2=4\] And radius becomes half \[{{r}_{1}}=r\]and \[{{r}_{2}}=\frac{1}{2}r\] Also, \[{{i}_{1}}={{i}_{2}}\] From the relation*n, the magnetic induction at the centre of a circular coil is given as \[B=\frac{{{\mu }_{0}}NI}{2r}\] Hence, \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{\frac{{{\mu }_{0}}{{N}_{1}}{{I}_{1}}}{2{{r}_{1}}}}{\frac{{{\mu }_{0}}{{N}_{2}}{{I}_{2}}}{2{{r}_{2}}}}\] ?(i) Putting, the given values in Eq. (i), we get \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{\mu }_{0}}\times 2\times {{I}_{1}}}{2r}\times \frac{2\times \frac{1}{2}r}{{{\mu }_{0}}\times 4\times {{I}_{1}}}=\frac{2}{8}\] \[\Rightarrow \] \[\frac{0.2}{{{B}_{2}}}=\frac{2}{8}\] \[{{B}_{2}}=\frac{8\times 0.2}{2}=0.8\,\]tesla
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