Question 329: Electromagnetic Induction

Question

The length of solenoid is $$I$$ whose windings are made of material of density $$D$$ and resistivity $$\rho$$. The winding resistance is $$R$$. The inductance of solenoid is [$$m=$$ mass of winding wire, $$\mu_0=$$ permeability of free space]

Accepted Answer

$$\begin{aligned}
&	ext { The self-inductance of solenoid, }\
&L=\mu_0 N^2 \frac{A}{l}\quad 	ext{.... (i)}
\end{aligned}$$
where, $$A$$ is the area of cross-section of solenoid and $$N$$ is number of turns.
If $$x$$ is the length of wire, then
$$R=\frac{ho x}{A} 	ext { and } m=A x D$$
$$\begin{aligned}
	herefore \quad R m & =\frac{ho x}{A}(A x D) \
\Rightarrow \quad x & =\sqrt{\frac{R m}{ho D}}\quad 	ext{.... (ii)}
\end{aligned}$$
$$	ext { Also, } x=2 \pi r N \Rightarrow N=\frac{x}{2 \pi r}=\frac{1}{2 \pi r} \sqrt{\frac{R m}{ho D}}\quad 	ext{[from Eq. (ii)]]}$$
$$\begin{aligned}
&	ext { Substituting values in Eq. (i), we get }\
&\begin{aligned}
L=\mu_0\left(\frac{\sqrt{R m}}{2 \pi r \sqrt{ho D}}ight)^2 	imes \frac{A}{l} & =\mu_0 \frac{R m}{4 \pi^2 r^2 ho D} \cdot \frac{\pi r^2}{l} \
& =\frac{\mu_0}{4 \pi I}\left(\frac{R m}{ho D}ight)
\end{aligned}
\end{aligned}$$

Answer

$$\frac{\mu_0}{4 \pi I}\left(\frac{R m}{\rho D}\right)$$

Answer

$$\frac{\mu_0}{2 \pi I}\left(\frac{R m}{\rho D}\right)$$

Answer

$$\frac{\mu_0}{2 \pi I}\left(\frac{\rho D}{R m}\right)$$

Answer

$$\frac{\mu_0}{4 \pi /}\left(\frac{\rho D}{R m}\right)$$

Electromagnetic Induction

The length of solenoid is $I whose windings are made of material of density D and resistivity \rho. The winding resistance is R. The inductance of solenoid is [m= mass of winding wire, \mu_0=$ permeability of free space]

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