Question

A solenoid $30cm$ long is made by winding $2000$ loops of wire on an iron rod whose cross-section is $1.5c{m}^2$. If the relative permeability of the iron is $6000$. What is the self inductance of the solenoid?

• A. $15H$
• B. $2.5H$
• C. $3.5H$
• D. $0.5H$

Answer 1

The correct option is A

$15H$

Step 1: Given data

Length of a solenoid, $L=30cm=0.3m$

Cross sectional area of the solenoid, $A=1.5c{m}^2=1.5\times {10}^{-4}{m}^2$

Number of loops = $2000$

Relative permeability of the iron ${\mu }_r=6000$

Let permeability of free space ${\mu }_0=4\mathrm{\pi }\times {10}^{-7}H/\mathrm{m}$

Step 2: Find self inductance of the solenoid

Permeability is the measure of magnetization that a material obtains in response to an applied magnetic field. A relative permeability is defined as the ratio of the material permeability to the permeability of free space (or vacuum)

${\mu }_r=\frac{\mu }{{\mu }_0}$

Where,

$\mu =$permeability of substance

$$\begin{gathered} \mu ={\mu }_0\times {\mu }_r \\ \mu =4\mathrm{\pi }\times {10}^{-7}\times 6000 \\ \mathrm{\mu }=75.42\times {10}^{-4}H/\mathrm{m} \end{gathered}$$

Self-inductance of a solenoid is defined as the flux associated with the solenoid when a unit amount of current is passed through it. It is given by

$$\begin{gathered} =\frac{{\mu }_0{\mu }_r{N}^{2}A}{L} \\ =\frac{4\mathrm{\pi }\times {10}^{-7}\times 6000\times {2000}^2\times 1.5\times {10}^{-4}}{0.3} \\ =15H \end{gathered}$$

Hence, option A is correct.