question_answer

Two circular coils X and Y have equal number of turn and carry equal currents in the same sense and subtend same solid angle at point O. If the smaller coil X is midway between O and Y , then if we represent the magnetic induction due to bigger coil Y at O as By and that due to smaller coil X at\[O\text{ }{{B}_{X}}\], then:

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 \[\therefore {{B}_{Y}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi I{{(2r)}^{2}}}{{{[{{(2r)}^{2}}+{{(d)}^{2}}]}^{3/2}}}\] \[{{B}_{X}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi I{{(r)}^{2}}}{{{\left[ {{r}^{2}}+{{\left( \frac{d}{2} \right)}^{2}} \right]}^{3/2}}}\] \[\therefore \frac{{{B}_{y}}}{{{B}_{X}}}=\frac{4}{{{(4{{r}^{2}}+{{d}^{2}})}^{3/2}}}\times {{\left[ \frac{4{{r}^{2}}+{{d}^{2}}}{4} \right]}^{3/2}}\] \[=\frac{4}{{{(4)}^{3/2}}}=\frac{4}{8}=\frac{1}{2}\]