question_answer

A circular current carrying coil has a radius R The distance from the centre of the coil off the axis of the coil, where the magnetic induction is I/8th of its  value at the centre of  the coil is

A)  \[\sqrt{3}R\]                    

B)        \[R/\sqrt{3}\]

C)  \[\left( \frac{2}{\sqrt{3}} \right)R\]        

D)        \[\frac{R}{2\sqrt{3}}\]

Correct Answer: A

Solution :

For a circular coil of radius a carrying a current \[i\], the magnetic field at point\[P\], distance\[x\]from coil is given by                 \[B=\frac{{{\mu }_{0}}i\,\,{{a}^{2}}}{2{{({{a}^{2}}+{{x}^{2}})}^{3/2}}}N{{A}^{-1}}{{m}^{-1}}\]       ? (i) At the centre of coil\[x=0\] \[\therefore \]  \[B=\frac{{{\mu }_{0}}i}{2a}N{{A}^{-1}}{{m}^{-1}}\]                         ? (ii) Given,   \[B=\frac{1}{8}B\] \[\therefore \]  \[\frac{{{\mu }_{0}}i\,\,{{a}^{2}}}{2{{({{a}^{2}}+{{x}^{2}})}^{3/2}}}=\frac{1}{8}\left( \frac{{{\mu }_{0}}i}{2a} \right)\] \[\Rightarrow \]               \[\frac{{{a}^{2}}}{{{({{a}^{2}}+{{x}^{2}})}^{3/2}}}=\frac{1}{8a}\] \[\Rightarrow \]               \[8\,\,{{a}^{3}}={{({{a}^{2}}+{{x}^{2}})}^{3/2}}\]                 \[\Rightarrow \]\[{{a}^{2}}+{{x}^{2}}=4{{a}^{2}}\] \[\Rightarrow \]               \[x=\sqrt{3\cdot a}\] Given,   \[a=R\]                 \[x=\sqrt{3}\,\,R\]