Question

The magnetic field inside a long solenoid of 50 turns cm−1 is increased from 2.5 × 10−3 T to 2.5 T when an iron core of cross-sectional area 4 cm2 is inserted into it. Find (a) the current in the solenoid (b) the magnetisation I of the core and (c) the pole strength developed in the core.

Answer 1

Explanation:

Step 1:

The strength of the magnetic field without iron core, $B_{1}=2.5 \times 10^{-3} T$

Magnetic field after iron core is inserted, $B_{2}=2.5 T$

Cross section area of iron core, $A=4 \times 10^{-4} \mathrm{m}^{2}$

Number of turns in length per segment, $\mathrm{n}=50 \text { turns } / \mathrm{cm}=5000 \text { turns/m }$

(a) Magnetic field produced with a solenoid

B is known as  

$B=\mu_{0} n i$  

where i = In solenoid, electrical current

Step 2:

On substituting the values, we get  

$2.5 \times 10^{-3}=4 \pi \times 10^{-7} \times 5000 \times \mathrm{i}$

$i=\frac{2.5 \times 10^{-3}}{4 \pi \times 10^{-7} \times 5000}$

$=0.398 A=0.4 A$

(b)  I is given by Magnetisation

$I=\frac{B}{\mu_{0}}-\mathrm{H}$

Where B is a net magnetic field after the core has been applied, that is, B=2.5 T.

And μ_0 H will be the magnetization field, that is to say the difference between the forces of the two magnetic fields.

$I=\frac{2.5 \times 10^{-3}}{4 \pi \times 10^{-7}} \quad . B 2-B 1$

$I=\frac{2.5\left(1-\frac{1}{1000}\right)}{4 \pi \times 10^{-7}}$

$I \approx 2 \times 10^{6} A / m$

(c) Magnetization Intensity I is given by,

$I=\frac{M}{V}$

$I=\frac{m \times 2 I}{A \times 2 I}=\frac{m}{A}$

$m=l A$  

$m=2 \times 10^{2} \times 4 \times 10^{-4}$

 $m=800 A-m$