Question

A solenoid having a core of ${10}^{-3}{m}^2$ in cross-section, half air (${\mu }_r=1$) and half iron (${\mu }_r=500$) is $1m$ long, the no. of turns are $2000$. What is coefficient of self induction?

• A. $0.5H$
• B. $0.75H$
• C. $1.26H$
• D. $2.25H$

Answer 1

The correct option is A $0.5H$

Given: A=${10}^{-3}$,${\mu }_r=1\left(air\right),{\mu }_r=500\left(iron\right),N=2000$

Solution: Self inductance formula :

$L_1=\frac{{\mu }_0{\mu }_r{N}^{2}A}{l}$

$L_2=\frac{{\mu }_0{\mu }_r{N}^{2}A}{l}$

$L=l_1+l_2=\frac{{\mu }_0{N}^{2}A}{l}[\frac{{\mu }_{r1}}{2}+\frac{{\mu }_{r2}}{2}]$

Remember here that $\frac{{\mu }_{r1}}{2}$ is taken because in solenoid half of the length is of air and half of iron.

$L=\frac{4\pi \times {10}^{-7}\times (2000{)}^2\times {10}^{-3}}{1}[\frac{1}{2}+\frac{500}{2}]$

L=1.26 H