Question

How much length of a very wire is required to obtain a solenoid of length $I_0$ and inductance L :

• A. $\sqrt{\frac{2\pi Ll}{{\mu }_0}}$
• B. $\sqrt{\frac{4\pi Ll}{{\mu }_0^2}}$
• C. $\sqrt{\frac{4\pi Ll}{{\mu }_0}}$
• D. $\sqrt{\frac{8\pi Ll}{{\mu }_0}}$

Answer 1

Answer: (C)

$\sqrt{\frac{4\pi Ll}{{\mu }_0}}$

For solenoid, if N is the number of turns, A is cross section area of solenoid, l is the length of wire then, $L\frac{{\mu }_0{N}^{2}A}{1}$

Where L is inductance of solenoid.

Cross section area if solenoid $=A=\pi r^2$ { where r is the radius of circular part}

length of wire used to form = circumference of circle $\times$ number of turns

$L_1=2\pi rN$

So, $L={\mu }_0{N}^2\pi r^2/I$

$={\mu }_0(2\pi Nr{)}^2/4\pi l$

$={\mu }_{0}L{{}_1}^2/4\pi l$

$4\pi Ll={\mu }_{0}L{{}_1}^2\Rightarrow L{{}_1}^2=\left(4\pi Ll/{\mu }_0\right)$

$L_1==\sqrt{\frac{4\pi Ll}{{\mu }_0}}$