question_answer

Two circular coils \[C\] and \[D\] have equal number of turns and carry equal currents in the same direction in the same sense and subtend same solid angle at point \[O\] as shown in figure. The smaller coil \[C\] is midway between \[O\] and\[D\]. If we represent magnetic field induction due to bigger coil and smaller coil \[C\] as \[{{B}_{D}}\] and \[{{B}_{C}}\] respectively, then By \[{{B}_{D}}/{{B}_{C}}\] is

A) \[1:4\]                                  

B) \[1:2\]

C) \[2:1\]                                  

D)  \[1:1\]

Correct Answer: B

Solution :

Since two coils subtend the same solid angle at \[O\], hence area of coil\[D=4\times \]area of coil\[C\] Therefore, radius of coil\[D=2\times \]radius of coil\[C\] \[\therefore \]  \[{{B}_{D}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi /{{(2r)}^{2}}}{{{[{{(2r)}^{2}}+{{a}^{2}}]}^{3/2}}}\] and        \[{{B}_{C}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi /{{(r)}^{2}}}{{{[{{r}^{2}}+{{(a/2)}^{2}}]}^{3/2}}}\] \[\therefore \]  \[\frac{{{B}_{D}}}{{{B}_{C}}}=\frac{4}{{{(4{{r}^{2}}+{{a}^{2}})}^{3/2}}}\times {{\left[ \frac{4{{r}^{2}}+{{a}^{2}}}{4} \right]}^{3/2}}\]                       \[=\frac{4}{8}=\frac{1}{2}\]