A) 3
B) 4
C) 6
D) 2
Correct Answer: B
Solution :
| [b] 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, |
| \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi {{i}_{2}}}{{{r}_{2}}}\] (i) |
| 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 \[{{r}_{1}}=2{{r}_{2}}\] |
| \[\therefore \] \[\frac{{{i}_{1}}}{2{{r}_{2}}}=\frac{{{i}_{2}}}{{{r}_{2}}}\] |
| 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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