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 \[{{B}_{D}}/{{B}_{C}}\]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 \[\therefore \] \[{{B}_{D}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi l{{(2r)}^{2}}}{{{[{{(2r)}^{2}}+{{a}^{2}}]}^{3/2}}}\] and \[{{B}_{C}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi l{{(r)}^{2}}}{{{[{{r}^{2}}+{{(a/2)}^{2}}]}^{3/2}}}\] \[\therefore \] \[\frac{{{B}_{D}}}{{{B}_{C}}}=\frac{4}{{{(4{{r}^{2}}+{{a}^{2}})}^{3/2}}}\times {{\left[ \frac{4{{r}^{2}}+{{a}^{2}}}{4} \right]}^{3/2}}\] \[=\frac{4}{8}=\frac{1}{2}\]