question_answer

A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is:     AIEEE Solevd Paper-2013

A) \[9.1\times {{10}^{-11}}weber\]              

B) \[6\times {{10}^{-11}}weber\]

C)                        \[3.3\times {{10}^{-11}}weber\]              

D) \[6.6\times {{10}^{-9}}weber\]

Correct Answer: A

Solution :

First we find the mutual inductance of the assembly. Let current I flow through the larger loop. The strength of induction at the smaller loop \[=\frac{{{\mu }_{0}}I{{R}^{2}}}{2{{\left( {{R}^{2}}+{{d}^{2}} \right)}^{3/2}}}\] \[\therefore \]Flux through smaller loop\[=\frac{{{\mu }_{0}}\pi I{{R}^{2}}{{r}^{2}}}{2{{\left( {{R}^{2}}+{{d}^{2}} \right)}^{3/2}}}\] \[\therefore \]Mutual inductance\[=\frac{{{\mu }_{0}}{{R}^{2}}{{r}^{2}}}{2{{\left( {{R}^{2}}+{{d}^{2}} \right)}^{3/2}}}\] \[\therefore \]Flux linked through coil\[=\frac{{{\mu }_{0}}\pi {{R}^{2}}{{r}^{2}}}{2{{\left( {{R}^{2}}+{{d}^{2}} \right)}^{3/2}}}.i\] \[=\frac{4\pi \times {{10}^{-7}}\times \pi \times 4\times {{10}^{-2}}\times 9\times {{10}^{-6}}\times 2}{2{{\left( 4\times {{10}^{-2}}+2.25\times {{10}^{-2}} \right)}^{3/2}}}\] \[=\frac{16}{2}\times \frac{10\times 18}{15.625}\times \left( {{10}^{+3-15}} \right)=9.1\times {{10}^{-11}}Weber\]