A) \[\frac{{{\mu }_{{}^\circ }}Rm}{4\pi R\rho D}\]
B) \[\frac{{{\mu }_{{}^\circ }}\ell m}{4\pi R\rho D}\]
C) \[\frac{{{\mu }_{{}^\circ }}}{4\pi \ell }\frac{{{R}^{2}}m}{\rho D}\]
D) \[\frac{{{\mu }_{{}^\circ }}}{2\pi \ell }\frac{\ell \,m}{\rho D}\]
Correct Answer: A
Solution :
[a] In a solenoid \[L=\frac{{{\mu }_{o}}{{N}^{2}}A}{\ell }\]if \[\times \]is the length of the wire and ?a? is cross section area \[R=\frac{\rho x}{a}m=a\times D\] \[Rm\,=\frac{\rho \,x\,a\,x\,D}{a}\] \[x=\sqrt{\frac{Rm}{\rho D}}\] \[Also\,x=2\pi N,\,N=\frac{x}{2\pi r}\,put\,this\,\left[ \frac{L\,={{\mu }_{o}}N_{\ell }^{2}A}{\ell } \right]\] We get \[L={{\mu }_{o}}{{\left[ \frac{x}{2\pi r} \right]}^{2}}\frac{\pi {{r}^{2}}}{t}=\frac{{{\mu }_{0}}}{4\pi \ell }\frac{Rm}{\rho D}\]
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