A) 4.6 mH
B) 6.9 mH
C) 2.3 mH
D) 9.2 mH
Correct Answer: C
Solution :
For a solenoid, \[B={{\mu }_{0}}ni\] where, \[n=\frac{N}{2\pi r}\] \[\Rightarrow \] \[B=\frac{{{\mu }_{0}}Ni}{2\pi r}\] Flux linked with the solenoid is \[\phi =NBA\] \[\Rightarrow \] \[\phi =\frac{{{\mu }_{0}}{{N}^{2}}iA}{2\pi r}\] Self-inductance of the toroid \[L=\frac{\phi }{i}=\frac{{{\mu }_{0}}{{N}^{2}}A}{2\pi r}\] \[=\frac{4\pi \times {{10}^{-7}}\times {{(1200)}^{2}}\times 12\times {{10}^{-4}}}{2\pi \times 15\times {{10}^{-2}}}\] \[L=2.3\times {{10}^{-3}}H=2.3\,mH\]
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