The length of solenoid is ' $\ell$ 'whose windings are made of material of density 'D' and resistivity 'Q'.…

The length of solenoid is ' $\ell$ 'whose windings are made of material of density 'D' and resistivity 'Q'. The winding resistance is 'R'. The inductance of solenoid is $\left(\mathrm{m}=\right.$ mass of winding wire, $\mu_{0}=$ permeability of free space)
  1. $\frac{\mu_{0}}{2 \pi \ell}\left(\frac{R m}{Q D}\right)$
  2. $\frac{\mu_{0}}{4 \pi \ell}\left(\frac{R m}{Q D}\right)$
  3. $\frac{\mu_{0}}{2 \pi \ell}\left(\frac{Q \mathrm{D}}{\mathrm{Rm}}\right)$
  4. $\frac{\mu_{0}}{4 \pi \ell}\left(\frac{\mathrm{Q} \mathrm{D}}{\mathrm{Rm}}\right)$

Solution

We know, $\mathrm{L}=\mu_{0} \mathrm{~N}^{2} \frac{\mathrm{A}}{\ell}$ If $\mathrm{x}$ is the length of the wire and $a$ is the area of cross section $\mathrm{R}=\frac{\rho \mathrm{x}}{\mathrm{a}} \quad \mathrm{m}=\mathrm{axD}$ $\mathrm{Rm}=\frac{\rho \mathrm{x}}{\mathrm{a}} \times \mathrm{axD} \quad \therefore \quad \mathrm{x}=\sqrt{\frac{\mathrm{Rm}}{\rho \mathrm{D}}}$ Also, $x=2 \pi r N$ $\therefore \quad N=\frac{X}{2 \pi}$ $\because$ $=\mu_{0}\left(\frac{\mathrm{x}}{2 \pi \mathrm{r}}\right)^{2} \frac{\pi \mathrm{r}^{2}}{\ell}=\frac{\mu_{0}}{4 \pi \ell} \frac{\mathrm{Rm}}{\mathrm{QD}}$

Asked in: MHT CET 2020 (16 Oct Shift 2)

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