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

Explanation:

Step 1:

Given data in the question  

Number of turns in length per segment, $n=40 \frac{\text { lums }}{c m}=4000 \frac{\text { turns }}{m}$

Area of Magnetization, $\mathrm{H}=4 \times 10^{4} \mathrm{A} / \mathrm{m}$

Step 2:

The magnetic field within a solenoid (B) is shown by,

$B=\mu_{0} n I$

Where,  

n = number of turns in length per unit.  

I =  by the solenoid current.

Step 3:

$\therefore \frac{B}{\mu_{0}}=n I=H$

To find H :                                          

$H=\frac{N}{l} I$

$I=\frac{H l}{N}=\frac{H}{n}$

$I=\frac{4 \times 10^{4}}{4000}$

$I=\frac{4 \times 10}{4}$

$=10 A$