A) \[L=\frac{{{\mu }_{0}}N{{r}^{2}}}{2R}\]
B) \[L=\frac{{{\mu }_{0}}Nr}{2R}\]
C) \[L=\frac{{{\mu }_{0}}N{{r}^{2}}}{R}\]
D) \[L=\frac{{{\mu }_{0}}{{N}^{2}}{{r}^{2}}}{2R}\]
Correct Answer: D
Solution :
The coefficient of self-inductance of the toroid \[L=\frac{Gf}{i},\] \[\phi =NAB\] and \[B={{\mu }_{0}}ni\] where, \[n=\frac{N}{2\pi R}\] \[\therefore \] \[\phi =N.\pi {{r}^{2}}\left( {{\mu }_{0}}\frac{N}{2\pi R}i \right)\] \[\phi =\frac{{{\mu }_{0}}{{N}^{2}}{{r}^{2}}i}{2R}\] \[L=\frac{\phi }{i}=\frac{{{\mu }_{0}}{{N}^{2}}{{r}^{2}}}{2R}\]
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