Question

The coercive force for a certain permanent magnet is 4.0 × 10⁴ A m⁻¹. This magnet is placed inside a long solenoid of 40 turns/cm and a current is passed in the solenoid to demagnetise it completely. Find the current.

Answer 1

Given:
Number of turns per unit length, n = 40 turns/cm = 4000 turns/m
Magnetising field, H = 4 × 10⁴ A/m
Magnetic field inside a solenoid (B) is given by,
B = µ₀nI,
where, n = number of turns per unit length.
I = current through the solenoid.
$\therefore \frac{B}{{\mu }_{\mathit{0}}}\mathit{}\mathit{=}\mathit{}nI\mathit{}\mathit{=}\mathit{}H$
$$\begin{gathered} \Rightarrow H\mathit{=}\frac{N}{l}I \\ \mathit{\Rightarrow }I\mathit{=}\frac{Hl}{N}\mathit{}\mathit{=}\mathit{}\frac{H}{n} \\ \Rightarrow I=\frac{4\times {10}^4}{4000}=10\mathrm{A} \end{gathered}$$