A) \[\frac{{{\mu }_{0}}}{4\pi l}.\frac{{{R}_{m}}}{\rho D}\]
B) \[\frac{{{\mu }_{0}}}{4\pi R}.\frac{{{l}_{m}}}{\rho D}\]
C) \[\frac{{{\mu }_{0}}}{4\pi l}.\frac{{{R}^{2}}m}{\rho D}\]
D) \[\frac{{{\mu }_{0}}}{2\pi R}.\frac{{{l}_{m}}}{\rho D}\]
Correct Answer: A
Solution :
For a solenoid \[L={{\mu }_{0}}{{N}^{2}}=\frac{A}{l}.2yx\]is the length of the wire and a is the area of cross-section, then \[R=\frac{\rho x}{a}\] and \[m=axD\] \[{{R}_{m}}=\rho \frac{x}{a}=axD\] \[x=\sqrt{\frac{{{R}_{m}}}{\rho D}}\] \[x=2\pi rN,N=\frac{x}{2\pi r}\] \[L={{\mu }_{0}}{{\left( \frac{{{x}^{2}}}{2\pi r} \right)}^{2}}\frac{\pi {{r}^{2}}}{l}\] \[=\frac{{{\mu }_{0}}}{4\pi l}.\frac{Rm}{\rho D}\] \[r=2cm=0.02\text{ }m\]
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