Question

A tightly wound solenoid of radius $‘a’$ and length $‘l’$ has $n$ turns per unit length. It carries an electric current $i$ . Magnetic field at a distance $\frac{l}{4}$ from one of the end (inside the solenoid on its axis) is $B=\frac{{\left(\mu \right)}_{0}ni\left(\sqrt{5}+3\right)}{\sqrt{K}}$ for $l=4a$ . Find the value of $K$ .

Answer 1

Answer: 40

Magnetic field inside the solenoid, $B=\frac{{\mu }_{0}ni}{2}\left[\sin \left({\theta }_1\right)+\sin \left({\theta }_2\right)\right]$
$\sin \left({\theta }_1\right)=\frac{3a}{\sqrt{(3a{)}^2+a^2}}=\frac{3}{\sqrt{10}}$
$\sin \left({\theta }_2\right)=\frac{a}{\sqrt{a^2+a^2}}=\frac{1}{\sqrt{2}}$

$\Rightarrow B=\frac{{\mu }_{0}ni}{2}\left[\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{10}}\right]$
$\Rightarrow B=\frac{{\mu }_{0}ni(\sqrt{5}+3)}{\sqrt{40}}$
$\therefore K=40$