A) \[\frac{{{B}_{Y}}}{{{B}_{X}}}=1\]
B) \[\frac{{{B}_{Y}}}{{{B}_{X}}}=2\]
C) \[\frac{{{B}_{Y}}}{{{B}_{X}}}=\frac{1}{2}\]
D) \[\frac{{{B}_{Y}}}{{{B}_{X}}}=\frac{1}{4}\]
Correct Answer: C
Solution :
[c] As two coils subtend the same solid angle at O, hence area of coil, b\[Y\text{ }=\text{ }4\text{ }\times \text{ }area\text{ }of\text{ }coil\text{ }X\] \[\left[ Solid\text{ }angle=\frac{area}{{{(\bot \,distance)}^{2}}} \right]\] i.e. radius of coil Y = 2 \[\times \] radius of coil X \[\therefore {{B}_{Y}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi I{{(2r)}^{2}}}{{{[{{(2r)}^{2}}+{{(d)}^{2}}]}^{3/2}}}\] \[{{B}_{X}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi I{{(r)}^{2}}}{{{\left[ {{r}^{2}}+{{\left( \frac{d}{2} \right)}^{2}} \right]}^{3/2}}}\] \[\therefore \frac{{{B}_{y}}}{{{B}_{X}}}=\frac{4}{{{(4{{r}^{2}}+{{d}^{2}})}^{3/2}}}\times {{\left[ \frac{4{{r}^{2}}+{{d}^{2}}}{4} \right]}^{3/2}}\] \[=\frac{4}{{{(4)}^{3/2}}}=\frac{4}{8}=\frac{1}{2}\]You need to login to perform this action.
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