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 $\left(B\right)$ is given by,
$B={\mathrm{\mu }}_{0}ni$,
where i = electric current in the solenoid
2.5 × 10⁻³ = 4$\mathrm{\pi }$ × 10⁻⁷ × 5000 × i
$$\begin{gathered} \Rightarrow i=\frac{2.5\times {10}^{-3}}{4\mathrm{\pi }\times {10}^{-7}\times 5000} \\ =0.398\mathrm{A}=0.4\mathrm{A} \end{gathered}$$
$$\begin{gathered} \left(b\right)\mathrm{Magneti}\text{sa}\mathrm{tion}\left(I\right)\mathrm{is}\mathrm{given}\mathrm{by} \\ I\mathit{=}\frac{B}{{\mu }_{\mathit{0}}}\mathit{-}H, \\ \text{w}\mathrm{here}B\mathrm{is}\mathrm{the}\mathrm{net}\mathrm{magnetic}\mathrm{field}\mathrm{after}\mathrm{introducing}\mathrm{the}\mathrm{core},\mathrm{i}.\mathrm{e}.B=2.5\mathrm{T}. \\ \mathrm{And}{\mathrm{\mu }}_{0}H\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{magnetising}\mathrm{field},\mathrm{i}.\mathrm{e}.\mathrm{the}\mathrm{diffrence}\mathrm{between}\mathrm{the}\mathrm{two}\mathrm{magnetic}\mathrm{fields}\text{'}\mathrm{strengths}. \\ \Rightarrow I=\frac{2.5\times {10}^{-3}}{4\mathrm{\pi }\times {10}^{-7}}.\left(B_2-B_1\right) \\ \Rightarrow I=\frac{2.5\left(1-{\displaystyle \frac{1}{1000}}\right)}{4\mathrm{\pi }\times {10}^{-7}} \\ \Rightarrow I\approx 2\times {10}^6\mathrm{A}/\mathrm{m} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{c}\right)\mathrm{Intensity}\mathrm{of}\mathrm{magneti}\text{s}\mathrm{ation}\left(I\right)\mathrm{is}\mathrm{given}\mathrm{by}, \\ I\mathit{}\mathit{=}\mathit{}\frac{M}{V} \\ \mathit{\Rightarrow }I\mathit{=}\mathit{}\frac{m\times 2I}{A\times 2I}\mathit{}\mathit{=}\mathit{}\frac{m}{A} \\ \mathit{\Rightarrow }m\mathit{}\mathit{=}\mathit{}lA \end{gathered}$$
$\Rightarrow$m= 2 × 10⁶ × 4 × 10⁻⁴
$\Rightarrow$m= 800 A-m