MHT CET · Physics · Magnetic Effects of Current
The length of solenoid is ' \(\ell\) 'whose windings are made of material of density 'D' and resistivity ' \(0^{\prime}\). The winding resistance is 'R'. The inductance of solenoid is \(\left(\mathrm{m}=\right.\) mass of winding wire, \(\mu_{0}=\) permeability of free space)
- A \(\frac{\mu_{0}}{2 \pi \ell}\left(\frac{R m}{Q D}\right)\)
- B \(\frac{\mu_{0}}{4 \pi \ell}\left(\frac{R m}{Q D}\right)\)
- C \(\frac{\mu_{0}}{2 \pi \ell}\left(\frac{\varrho \mathrm{D}}{\mathrm{Rm}}\right)\)
- D \(\frac{\mu_{0}}{4 \pi \ell}\left(\frac{\mathrm{Q} \mathrm{D}}{\mathrm{Rm}}\right)\)
Answer & Solution
Correct Answer
(B) \(\frac{\mu_{0}}{4 \pi \ell}\left(\frac{R m}{Q D}\right)\)
Step-by-step Solution
Detailed explanation
(D)
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, \(x=2 \pi r N\) \(\therefore \quad N=\frac{X}{2 \pi}\)
\(\because\) \(=\mu_{0}\left(\frac{\mathrm{x}}{2 \pi \mathrm{r}}\right)^{2} \frac{\pi \mathrm{r}^{2}}{\ell}=\frac{\mu_{0}}{4 \pi \ell} \frac{\mathrm{Rm}}{\mathrm{pD}}\)
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, \(x=2 \pi r N\) \(\therefore \quad N=\frac{X}{2 \pi}\)
\(\because\) \(=\mu_{0}\left(\frac{\mathrm{x}}{2 \pi \mathrm{r}}\right)^{2} \frac{\pi \mathrm{r}^{2}}{\ell}=\frac{\mu_{0}}{4 \pi \ell} \frac{\mathrm{Rm}}{\mathrm{pD}}\)
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