Question

Determine the flux density inside a bar of aluminium that is positioned within a solenoid of wire $1.2\text{m}$ long and having $600$ turns, carrying a current of $8\text{A}$. Magnetic susceptibility of aluminium is $2.5\times {10}^{-4}$ .

• A. $50.28\times {10}^{-2}\text{T}$
• B. $50.28\times {10}^{-3}\text{T}$
• C. $50.28\times {10}^{-7}\text{T}$
• D. $50.28\times {10}^{-4}\text{T}$

Answer 1

The correct option is D $50.28\times {10}^{-4}\text{T}$

We know that, the magnetic field intensity is given by,

$H=\frac{Ni}{l}=\frac{600\times 8}{1.2}=4000\text{A/m}$

Since, relative permeability, ${\mu }_r=\chi +1=2.5\times {10}^{-4}+1=1.00025$

So, $\mu ={\mu }_r{\mu }_0=1.00025\times 4\pi \times {10}^{-7}=12.57\times {10}^{-7}\text{H/m}$

Therfore, $B=\mu H=12.57\times {10}^{-7}\times 4000=50.28\times {10}^{-4}\text{T}$

So, option (d) is the right answer.