A) 3
B) 4
C) 6
D) 2
Correct Answer: B
Solution :
Magnetic field at the centre of a circular coil is \[B=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi i}{r}\] Where\[i\]is current flowing in the coil and r is radius of coil. At the centre of coil 1, \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi {{i}_{1}}}{{{r}_{1}}}\] (i) At the centre of coil 2, \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi {{i}_{2}}}{{{r}_{2}}}\] (ii) But \[{{B}_{1}}={{B}_{2}}\] \[\therefore \]\[\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi {{i}_{1}}}{{{r}_{1}}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi {{i}_{2}}}{{{r}_{2}}}\]or \[\frac{{{i}_{1}}}{{{r}_{1}}}=\frac{{{i}_{2}}}{{{r}_{2}}}\] As \[{{r}_{1}}=2{{r}_{2}}\] \[\therefore \] \[\frac{i}{2{{r}_{2}}}=\frac{{{i}_{2}}}{{{r}_{2}}}\]or \[{{i}_{1}}=2{{i}_{2}}\] (iii) Now, ratio of potential differences \[\frac{{{V}_{2}}}{{{V}_{1}}}=\frac{{{i}_{2}}\times {{r}_{2}}}{{{i}_{1}}\times {{r}_{1}}}=\frac{{{i}_{2}}\times {{r}_{2}}}{2{{i}_{2}}\times 2{{r}_{2}}}=\frac{1}{4}\] \[\therefore \] \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{4}{1}\]
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