Question

The coercive force for a certain permanent magnet is $4\times {10}^{4}A{m}^{-1}$ This magnet is placed inside a long solenoid of 40turns/cm and a current is passed in the solenoid to demagnetise it completely. Find the current.

Answer 1

Hint
Here, we use the formula of magnetic field of the solenoid having N turns per unit length and current I is passing through it i.e. $B={\mu }_{0}NI$, as the value of magnetising field is given i.e. $H=4\times {10}^{4}A{m}^{-1}$and we also know that $H={\displaystyle \frac{B}{{\mu }_0}}$, on rearranging the terms and substituting the values we get the value of current I.

Complete step by step answer
It is given that, Number of turns per unit length is 40 turns/cm = 4000 turns/m
The cohesive force or magnetising field is $H=4\times {10}^{4}A{m}^{-1}$
As we know that the magnetic field inside a solenoid is given by
$B={\mu }_{0}NI$ ……………………… (1)
Where, B is the magnetic field
N is the number of turns per unit length
I is the current passing through the solenoid
As we also know the relation between the magnetising field and magnetic field is
$\Rightarrow H={\displaystyle \frac{B}{{\mu }_0}}$……………………… (2)
From equation (1) and (2), we get
$\Rightarrow H=NI$
On substituting the values, we get
$\Rightarrow 4\times {10}^4=4000\times I$
$\Rightarrow I=10A$
Hence, 10A is the current required to demagnetise the magnet completely.

Note
Here it must be noticed that B is the magnetic flux density and H is the magnetic intensity. Here, electric current I produces around itself the magnetic field strength H regardless the type of the surrounding medium. Magnetic flux density B is the response of the medium to the applied