Question

A solenoid has a core of a material with relative permeability 501 . The windings of the solenoid are insulated from the core find carry a current of $2.5 A$. If the number of turns are 900 per metre. The magnetization in $A / m$ is

• A. $1.125 \times 10^6$
• B. $2.8 \times 10^6$
• C. $2.25 \times 10^6$
• D. $1.69 \times 10^6$

Answer 1

Answer: (A)

Given, relative permeability, $\mu_r=501$
Current through solenoid, $I=2.5 A$
and number of turns per unit length, $n=\frac{N}{l}=900$
As we know that, $B=\mu_0 H$
$\Rightarrow H=\frac{B}{\mu_0}$
where,
$B=$ magnetic flux density / magnetic induction,
$\mu_0=$ free space permeability
and $H=$ magnetising field intensity.
Since, magnetic field due to solenoid, $B=\mu_0 n I$
Here, $\mu$ is permeability of medium
and $\mu=\mu_r \mu_0$
where, $\mu_r$ is relative permeability.
$ \therefore H=\frac{B}{\mu_0}=\frac{n \mu I}{\mu_0} $
$ =\frac{n \mu_r \mu_0 I}{\mu_0} $
$ \Rightarrow H=n \mu_r I$
$ =900 \times 501 \times 2.5 $
$ =1.12 \times 10^6 A / m$