Question 292: Electromagnetic Induction

Question

To manufacture a solenoid of length $$1 \mathrm{~m}$$ and inductance $$1 \mathrm{~mH}$$, the length of thin wire required is
(cross - sectional diameter of a solenoid is considerably less than the length)

Accepted Answer

Inductance of solenoid, $$\mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}$$ .....(i)
where, $$l=$$ length of solenoid.
$$\mathrm{A}=\pi \mathrm{r}^2=	ext { area of solenoid }$$
Let '$$x$$' be length of wire required.
$$\begin{aligned}
	herefore \quad \mathrm{x} & =	ext { circumference of solenoid } 	imes 	ext { no. of turns } \
& =2 \pi \mathrm{rN} . \
	herefore \quad \mathrm{N} & =\frac{\mathrm{x}}{2 \pi \mathrm{r}}
\end{aligned}$$
Substituting in equation (i),
$$\begin{aligned}
& \mathrm{L}=\frac{\mu_0\left(\frac{\mathrm{x}^2}{4 \pi^2 \mathrm{r}^2}ight) 	imes \pi \mathrm{r}^2}{l} \
& 	herefore \quad \mathrm{L}=\frac{\mu_0 \mathrm{x}^2}{4 \pi l} \
& 	herefore \quad \mathrm{x}^2=\frac{4 \pi \mathrm{L} l}{\mu_0} \
& 	herefore \quad \mathrm{x}=\sqrt{\frac{4 \pi \mathrm{L} l}{\mu_0}} \
\end{aligned}$$
Substituting the values,
$$\begin{aligned}
& x=\sqrt{\frac{4 	imes \pi 	imes 10^{-3} 	imes 1}{4 \pi 	imes 10^{-7}}} \
& x=\sqrt{10^4} \mathrm{~m} \
& x=0.10 \mathrm{~km}
\end{aligned}$$

Answer

0.10 m

Answer

0.10 km

Answer

1 km

Answer

10 km

Electromagnetic Induction

To manufacture a solenoid of length $1 \mathrm{~m} and inductance 1 \mathrm{~mH}$, the length of thin wire required is (cross - sectional diameter of a solenoid is considerably less than the length)

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