The magnetic field inside a long solenoid having 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 cm 2 is inserted into it. Find (a) the current in the solenoid, (b) the magnetization I of the core and (c) the pole strength developed in the core.

Question

Philoid · Accepted Answer

Given:

No. of turns/cm=50

Magnetic field inside without iron core=2.5× 10^-3T

Magnetic field inside solenoid with iron core=2.5T

Cross-sectional area of solenoid=4cm²

We know that magnetic field(B) inside the solenoid without iron core is given by

Where,

n=no. of turns in a given length

I=current through solenoid

Putting the values of B, I and μ₀

A

Therefore, current through the solenoid is given by 0.4A

Magnetization I of the core is given by formula

Where H=magnetization intensity

Let the magnetic field with iron core and without iron is given by B₂ and B₁ respectively.

Then,

Putting the values of B₁,B₂ and μ

Am^-1

Therefore, the magnetization in the core is 2× 10⁶A/m

Intensity of magnetization is given by the formula

Where,

M=magnetic moment given by the formula

…(i)

Where,

m=magnetic pole strength

l=length between two poles

also volume inside solenoid is given by

..(ii)

Using eqns.(i) and (ii) we get

Putting the values of I and A

A-m

Therefore, magnetic pole strength developed inside the core is given by 800A-m

The magnetic field inside a long solenoid having 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 magnetization I of the core and(c) the pole strength developed in the core.