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 :
[a] 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\[=\frac{{{\mu }_{0}}\pi {{R}^{2}}{{r}^{2}}}{2{{\left( {{R}^{2}}+{{d}^{2}} \right)}^{3/2}}}.i\] |
\[=\frac{4\pi \times {{10}^{-7}}\times \pi \times 4\times {{10}^{-2}}\times 9\times {{10}^{-6}}\times 2}{2{{\left( 4\times {{10}^{-2}}+2.25\times {{10}^{-2}} \right)}^{3/2}}}\] |
\[=\frac{16}{2}\times \frac{10\times 18}{15.625}\times \left( {{10}^{+3-15}} \right)=9.1\times {{10}^{-11}}Weber\] |
You need to login to perform this action.
You will be redirected in
3 sec