Question

An ideal solenoid having $5000turns/m$ has an aluminium core and carries a current of $5A$. If ${\chi }_{Al}=2.3\times {10}^{-5}$, then the magnetic field developed at center will be

• A. $0.031T$
• B. $0.048T$
• C. $0.027T$
• D. $0.050T$

Answer 1

The correct option is A $0.031T$

Given, $n=5000turns/m$, $i=5A$, ${\chi }_{Al}=2.3\times {10}^{-5}$, $B=$?
As, $B={\mu }_0(H+l)$
where, $H=\frac{B_0}{{\mu }_0}=\frac{{\mu }_{0}ni}{{\mu }_0}=ni$
$=5000\times 5=2.5\times {10}^{4}A/m$
and $l=\chi H=2.3\times {10}^{-5}\times 2.5\times {10}^4$
$=0.575A/m$
$\therefore B={\mu }_0(H+l)$
$=4\pi \times {10}^{-7}(2.5\times {10}^4+0.575)T$
$=0.031T$