A) a quarter of its first value
B) unaltered
C) fourtimes of its first value
D) half of its first value
Correct Answer: C
Solution :
As, magnetic field due to a circular coil axis centre is given by \[{{B}_{0}}=\frac{{{\mu }_{0}}nI}{2r}\] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{n}_{1}}}{{{n}_{2}}}\left( \frac{{{r}_{2}}}{{{n}_{1}}} \right)\] Where, \[{{n}_{1}}=1,\,\,{{n}_{2}}=2,\,\,{{r}_{2}}=\frac{{{r}_{1}}}{2}\] Then, \[\frac{{{B}_{2}}}{{{B}_{1}}}=\frac{4}{1}\] \[\Rightarrow \] \[{{B}_{2}}=4{{B}_{1}}\]You need to login to perform this action.
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