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 \[\phi =NBA\] \[\Rightarrow \] \[\phi =\frac{{{\mu }_{0}}{{N}^{2}}A}{2\pi r}\] \[\Rightarrow \] \[L=\frac{\phi }{I}=\frac{{{\mu }_{0}}{{N}^{2}}A}{2\pi r}\] \[\Rightarrow \] \[L=2.3\times {{10}^{-3}}H\]
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