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 $2.5A$. 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, $I=2.5A$
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, $B={\mu }_{0}nI$
Here, $\mu$ is permeability of medium
and $\mu ={\mu }_r{\mu }_0$
where, ${\mu }_r$ is relative permeability.
$\therefore H=\frac{B}{{\mu }_0}=\frac{n\mu I}{{\mu }_0}$
$=\frac{n{\mu }_r{\mu }_{0}I}{{\mu }_0}$
$\Rightarrow H=n{\mu }_{r}I$
$=900\times 501\times 2.5$
$=1.12\times 10^{6}A/m$