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

Given:-

Magnetic field strength without iron core, B₁ = 2.5 × 10⁻³ T

Magnetic field after introducing the iron core, B₂ = 2.5 T

Area of cross-section of the iron core, A = 4 × 10⁻⁴ m²

Number of turns per unit length, n = 50 turns/cm = 5000 turns/m

(a) Magnetic field produced by a solenoid (B) is given by,

\[B   =  \mu_0 ni\]

where i = electric current in the solenoid

2.5 × 10⁻³ = 4π × 10⁻⁷ × 5000 × i

\[\Rightarrow i = \frac{2 . 5 \times {10}^{- 3}}{4\pi \times {10}^{- 7} \times 5000}\]

\[ = 0 . 398  A = 0 . 4  A\]

 

(b) Magnetisation (I) is given by,

`I=B/mu_0-H,`

where B is the net magnetic field after introducing the core, i.e. B = 2.5 T.

And `mu_0H` will be the magnetising field, i.e. the difference between the two magnetic field's strengths.

\[ \Rightarrow I = \frac{2 . 5 \times {10}^{- 3}}{4\pi \times {10}^{- 7}} . \left( B_2 - B_1 \right)\] 

\[ \Rightarrow I = \frac{2 . 5\left( 1 - \frac{1}{1000} \right)}{4\pi \times {10}^{- 7}}\] 

\[ \Rightarrow I \approx 2 \times  {10}^6   A/m\]

(c) Intensity of magnetisation (I) is given by,

\[ I   =   \frac{M}{V}\]

\[ \Rightarrow I =   \frac{m \times 2I}{A \times 2I}   =   \frac{m}{A}\]

\[ \Rightarrow m   =   lA\]

\[\Rightarrow m=2\times10^2\times4\times10^{-4}\]

⇒ m = 800 A-m