A) \[\frac{{{\mu }_{0}}}{4\pi l}.\frac{Rm}{\rho D}\]
B) \[\frac{{{\mu }_{0}}}{4\pi R}.\frac{lm}{\rho D}\]
C) \[\frac{{{\mu }_{0}}}{4\pi l}.\frac{{{R}^{2}}m}{\rho D}\]
D) \[\frac{{{\mu }_{0}}}{2\pi R}.\frac{lm}{\rho D}\]
Correct Answer: A
Solution :
For a solenoid\[L={{\mu }_{0}}{{N}^{2}}\frac{A}{l}.\]If\[x\]is the length of the wire and a is the area of cross-section, then \[R=\frac{\rho x}{a}\]and \[m=axD\] \[Rm=\frac{\rho x}{a}.axD,\] \[\therefore \] \[x=\sqrt{\frac{Rm}{\rho D}}\] Also, \[x=2\pi rN,N=\frac{x}{2\pi r}\] \[\left( \because L=\frac{{{\mu }_{0}}{{N}^{2}}A}{l} \right)\] \[\therefore \] \[L={{\mu }_{0}}{{\left( \frac{x}{2\pi r} \right)}^{2}}.\frac{\pi {{r}^{2}}}{l}=\frac{{{\mu }_{0}}}{4\pi l}.\frac{Rm}{\rho D}\]
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